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Standard XII

Chemistry

Introduction to Alkanes

Q. 1 mole of ...

Question

Q. 1 mole of CH4 on combustion gave 33g CO2 thus % yield is

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Solution

CH4+2O2=CO2 +H2O
theoretically,
1 mol CH4 gives 1 mol CO2, that is 64g CO2.
but given,
actual yield = 33g
Therefore,
%yield=( actual yield/ theoretical yield) ×100
% yield=(33/64)×100
=51.56%

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2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O

Calculate the % yield of the combustion reaction.

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58 g

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2 C_{4} H_{10} + 13 O_{2} \to 8 C O_{2} + 10 H_{2} O

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0.30 g of an organic compound C, H and Oxygen on combustion yields 0.44g $C O_{2}$ and 0.18 g $H_{2} O$ . If one mole of compound weighs 60, then molecular formula of the compound is

Q. An organic compound contains

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Q. 1 mole of CH4 on combustion gave 33g CO2 thus % yield is

CH4+2O2=CO2 +H2O theoretically, 1 mol CH4 gives 1 mol CO2, that is 64g CO2. but given, actual yield = 33g Therefore, %yield=( actual yield/ theoretical yield) ×100 % yield=(33/64)×100 =51.56%

CH4+2O2=CO2 +H2O
theoretically,
1 mol CH4 gives 1 mol CO2, that is 64g CO2.
but given,
actual yield = 33g
Therefore,
%yield=( actual yield/ theoretical yield) ×100
% yield=(33/64)×100
=51.56%