When 22.4 L of H2 (g) is mixed with 11.2 L of Cl2 (g), each of at STP, the moles of HCl (g) formed is equal to from Chemistry Some Basic Concepts of Chemistry Class 12 Nagaland Board
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When 22.4 L of H2 (g) is mixed with 11.2 L of Cl2 (g), each of at STP, the moles of HCl (g) formed is equal to

  • 1 mole of HCl (g)

  • 2 moles of HCl (g)

  • 0.5 mole of (g) 

  • 1.5 mole of HCl(g)


A.

1 mole of HCl (g)

The given problem is related to the concept of the stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then identify the limiting reagent [possessing minimum number of moles and completely used up in the reaction]
The limiting reagent gives the moles of product formed in the reaction.

space space space space space space space space space space space space space space space space space space space space straight H subscript 2 space left parenthesis straight g right parenthesis space plus space space Cl subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow 2 HCl space left parenthesis straight g right parenthesis
initial space vol. space space 22.4 space straight L space space space space space space space space 11.2 space straight L space space space space space space 2 space mol
therefore space 22.4 space straight L space volume space at space STP space is space occupied space by
Cl subscript 2 space equals space 1 space mole
therefore space 11.2 space straight L space volume space will space be space occupied space by comma
Cl subscript 2 space equals space fraction numerator 1 space straight x space 11.2 over denominator 22.4 end fraction space mole
space equals space 0.5 space mol
Thus comma stack space straight H subscript 2 space with 1 mole below left parenthesis straight g right parenthesis space plus stack Cl subscript 2 with 0.5 below space left parenthesis straight g right parenthesis space rightwards arrow 2 HCl space left parenthesis straight g right parenthesis
Since comma Cl subscript 2 space possesses space minimum space number space of space moles comma
thus space it space is space the space limiting space reagent.
As space per space equation comma
1 space mol space Cl subscript 2 space identical to space 2 space mol space HCl
therefore space 0.5 space mol space Cl subscript 2 space equals space 2 space straight x 0.5 space mol space HCl
Hence comma space 1.0 space mole space of space HCl space left parenthesis straight g right parenthesis space is space priduced space by space 0.5 space mole space of space Cl subscript 2.

The given problem is related to the concept of the stoichiometry of chemical equations. Thus, we have to convert the given volumes into their moles and then identify the limiting reagent [possessing minimum number of moles and completely used up in the reaction]
The limiting reagent gives the moles of product formed in the reaction.

space space space space space space space space space space space space space space space space space space space space straight H subscript 2 space left parenthesis straight g right parenthesis space plus space space Cl subscript 2 space left parenthesis straight g right parenthesis space rightwards arrow 2 HCl space left parenthesis straight g right parenthesis
initial space vol. space space 22.4 space straight L space space space space space space space space 11.2 space straight L space space space space space space 2 space mol
therefore space 22.4 space straight L space volume space at space STP space is space occupied space by
Cl subscript 2 space equals space 1 space mole
therefore space 11.2 space straight L space volume space will space be space occupied space by comma
Cl subscript 2 space equals space fraction numerator 1 space straight x space 11.2 over denominator 22.4 end fraction space mole
space equals space 0.5 space mol
Thus comma stack space straight H subscript 2 space with 1 mole below left parenthesis straight g right parenthesis space plus stack Cl subscript 2 with 0.5 below space left parenthesis straight g right parenthesis space rightwards arrow 2 HCl space left parenthesis straight g right parenthesis
Since comma Cl subscript 2 space possesses space minimum space number space of space moles comma
thus space it space is space the space limiting space reagent.
As space per space equation comma
1 space mol space Cl subscript 2 space identical to space 2 space mol space HCl
therefore space 0.5 space mol space Cl subscript 2 space equals space 2 space straight x 0.5 space mol space HCl
Hence comma space 1.0 space mole space of space HCl space left parenthesis straight g right parenthesis space is space priduced space by space 0.5 space mole space of space Cl subscript 2.

3178 Views

Equal masses of H2, O2 and methane have been taken in a container of volume V at temperature 270 C in identical conditions. The ratio of the volumes of gases H2 : O2: CH4 would be 

  • 8:16:1

  • 16:8:1

  • 16:1:2

  • 8:1:2


C.

16:1:2

According to Avogadro's hypothesis,
Volume of a gas (V) straight alpha number of moles (n)
Therefore, the ratio of the volumes of gases can be determined in terms of their moles. The ratio of volumes of H2:O2: methane (CH4) is given by

straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space equals space straight n subscript straight H subscript 12 end subscript space colon straight n subscript straight O subscript 2 end subscript space colon space straight n subscript CH subscript 4 end subscript

rightwards double arrow space straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space equals space straight m subscript straight H subscript 2 end subscript over straight M subscript straight H subscript 2 end subscript colon space straight m subscript straight O subscript 2 end subscript over straight M subscript straight O subscript 2 end subscript space colon straight m subscript CH subscript 4 end subscript over straight M subscript CH subscript 4 end subscript
But space straight m subscript straight H subscript 2 end subscript space equals space straight m subscript straight O subscript 2 end subscript space equals space straight m subscript CH subscript 4 end subscript space equals space straight m space open square brackets therefore space straight n space equals space fraction numerator mass over denominator molar space massd end fraction close square brackets
Thus comma space space straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space space equals space straight m over 2 space equals space straight m over 1 space equals straight m over 16 space equals space 16 colon space 1 colon 2

According to Avogadro's hypothesis,
Volume of a gas (V) straight alpha number of moles (n)
Therefore, the ratio of the volumes of gases can be determined in terms of their moles. The ratio of volumes of H2:O2: methane (CH4) is given by

straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space equals space straight n subscript straight H subscript 12 end subscript space colon straight n subscript straight O subscript 2 end subscript space colon space straight n subscript CH subscript 4 end subscript

rightwards double arrow space straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space equals space straight m subscript straight H subscript 2 end subscript over straight M subscript straight H subscript 2 end subscript colon space straight m subscript straight O subscript 2 end subscript over straight M subscript straight O subscript 2 end subscript space colon straight m subscript CH subscript 4 end subscript over straight M subscript CH subscript 4 end subscript
But space straight m subscript straight H subscript 2 end subscript space equals space straight m subscript straight O subscript 2 end subscript space equals space straight m subscript CH subscript 4 end subscript space equals space straight m space open square brackets therefore space straight n space equals space fraction numerator mass over denominator molar space massd end fraction close square brackets
Thus comma space space straight V subscript straight H subscript 2 end subscript space colon space straight V subscript straight O subscript 2 end subscript space colon space straight V subscript CH subscript 4 end subscript space space equals space straight m over 2 space equals space straight m over 1 space equals straight m over 16 space equals space 16 colon space 1 colon 2

2981 Views

6.02 x 1020 molecules of urea are present in 100 mL of its solution. The concentration of solution is

  • 0.02 M

  • 0.01 M

  • 0.001 M

  • 0.1 M


B.

0.01 M

Given, number of molecules of urea =6.02 x 1020
therefore ,Number of moles
fraction numerator 6.02 space straight x space 10 to the power of 20 over denominator straight N subscript straight A end fraction
equals space fraction numerator 6.02 space x space 10 to the power of 20 over denominator 6.02 space x space space 10 to the power of 23 end fraction space equals space 1 space x 10 to the power of negative 3 end exponent space m o l
v o l u m e space o f space t h e space s o l u t i o n
space equals space 100 space m L space equals space 100 over 1000 L space equals space 0.1 space L
C o n c e n t r a t i o n space o f space u r e a space s o l u t i o n
left parenthesis i n space m o l space L to the power of negative 1 end exponent right parenthesis space equals space fraction numerator 1 space x space 10 to the power of negative 3 end exponent over denominator 0.1 end fraction space equals space m o l space L to the power of negative 1 end exponent

Given, number of molecules of urea =6.02 x 1020
therefore ,Number of moles
fraction numerator 6.02 space straight x space 10 to the power of 20 over denominator straight N subscript straight A end fraction
equals space fraction numerator 6.02 space x space 10 to the power of 20 over denominator 6.02 space x space space 10 to the power of 23 end fraction space equals space 1 space x 10 to the power of negative 3 end exponent space m o l
v o l u m e space o f space t h e space s o l u t i o n
space equals space 100 space m L space equals space 100 over 1000 L space equals space 0.1 space L
C o n c e n t r a t i o n space o f space u r e a space s o l u t i o n
left parenthesis i n space m o l space L to the power of negative 1 end exponent right parenthesis space equals space fraction numerator 1 space x space 10 to the power of negative 3 end exponent over denominator 0.1 end fraction space equals space m o l space L to the power of negative 1 end exponent

1596 Views

1.0 g of magnesium is burnt with 0.56 g of O2 in a closed vessel. Which reactant is left in excess and how much?

  • Mg, 0.16 g

  • O2, 0.16 g

  • Mg, 0.44 g

  • O2, 0.28 g


A.

Mg, 0.16 g

The balanced chemical equation is 
Mg + 1/2O2 --> MgO
24g    16 g        40g

From the above equation, it is clear that,
24 f Mg reacts with 16 g O2
thus, 1.0 g Mg reacts with 

16 over 24 space straight x space 0.67 space straight g space straight O subscript 2 space equals space 0.67 space straight g space straight O subscript 2
But space only space 0.56 space straight g space straight O subscript 2 space is space available space which space is space less space than space 0.67 space straight g. space
Thus comma space straight O subscript 2 space si space the space limiting space reagent.
further comma space 16 space straight g space straight O subscript 2 space reacts space with space Mg space 24 space straight g
therefore comma space 0.56 space straight g space will space reacts space with space Mg
equals space 24 over 16 space straight x space 0.56 space equals space 0.84 space straight g
therefore comma space Amount space of space Mg space left space unreacted space
equals space 1.0 minus 0.84 space straight g space Mg space equals space 0.16 space straight g space Mg
hence space comma space Mg space is space present space in space excess space and space 0.16 space straight g space Mg space is space left space behind space unreacted.

The balanced chemical equation is 
Mg + 1/2O2 --> MgO
24g    16 g        40g

From the above equation, it is clear that,
24 f Mg reacts with 16 g O2
thus, 1.0 g Mg reacts with 

16 over 24 space straight x space 0.67 space straight g space straight O subscript 2 space equals space 0.67 space straight g space straight O subscript 2
But space only space 0.56 space straight g space straight O subscript 2 space is space available space which space is space less space than space 0.67 space straight g. space
Thus comma space straight O subscript 2 space si space the space limiting space reagent.
further comma space 16 space straight g space straight O subscript 2 space reacts space with space Mg space 24 space straight g
therefore comma space 0.56 space straight g space will space reacts space with space Mg
equals space 24 over 16 space straight x space 0.56 space equals space 0.84 space straight g
therefore comma space Amount space of space Mg space left space unreacted space
equals space 1.0 minus 0.84 space straight g space Mg space equals space 0.16 space straight g space Mg
hence space comma space Mg space is space present space in space excess space and space 0.16 space straight g space Mg space is space left space behind space unreacted.

2123 Views

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

  • 1/4

  • 3/8

  • 1/2

  • 1/8


D.

1/8

we have given, 
a number of moles of hydrogen andleft parenthesis straight n subscript straight H subscript 2 end subscript right parenthesis that of oxygen areleft parenthesis straight n subscript straight O subscript 2 end subscript right parenthesis not equal.
therefore space We space have comma space the space relation space between space ratio space of space number space of space moles space
escaped space and space ratio space of space molecular space mass.
straight n subscript straight O subscript 2 end subscript over straight n subscript straight H subscript 2 end subscript space equals space square root of straight M subscript straight H subscript 2 end subscript over straight M subscript straight O subscript 2 end subscript end root


where comma space straight M space equals space molecular space mass space of space the space molecules.
straight n subscript straight O subscript 2 end subscript over straight n subscript straight H subscript 2 end subscript space equals space square root of 2 over 32 end root
straight n subscript straight O subscript 2 end subscript over straight n subscript straight H subscript 2 end subscript space equals space square root of 1 over 16 end root

fraction numerator straight n subscript straight O subscript 2 end subscript over denominator 0.5 end fraction space equals space 1 fourth

straight n subscript straight O subscript 2 end subscript space equals fraction numerator 0.5 over denominator 4 end fraction space equals 1 divided by 8

we have given, 
a number of moles of hydrogen andleft parenthesis straight n subscript straight H subscript 2 end subscript right parenthesis that of oxygen areleft parenthesis straight n subscript straight O subscript 2 end subscript right parenthesis not equal.
therefore space We space have comma space the space relation space between space ratio space of space number space of space moles space
escaped space and space ratio space of space molecular space mass.
straight n subscript straight O subscript 2 end subscript over straight n subscript straight H subscript 2 end subscript space equals space square root of straight M subscript straight H subscript 2 end subscript over straight M subscript straight O subscript 2 end subscript end root


where comma space straight M space equals space molecular space mass space of space the space molecules.
straight n subscript straight O subscript 2 end subscript over straight n subscript straight H subscript 2 end subscript space equals space square root of 2 over 32 end root
straight n subscript straight O subscript 2 end subscript over straight n subscript straight H subscript 2 end subscript space equals space square root of 1 over 16 end root

fraction numerator straight n subscript straight O subscript 2 end subscript over denominator 0.5 end fraction space equals space 1 fourth

straight n subscript straight O subscript 2 end subscript space equals fraction numerator 0.5 over denominator 4 end fraction space equals 1 divided by 8

3652 Views