Question

When the following reaction was carried out, the student obtained 5 g of CO₂. If the student started with 10 g of CH₄ and 10 g of O₂, what is the % yield of the reaction?

CH₄ + 2 O₂ --> CO₂ + 2 H₂O

Answer 1

sol:-

• atomic wt of H = 1 g/mol
• atomic wt. of O = 16 g/mol.
• atomic wt. of C = 12 g/mol.
• molecular wt. of CO2 = (12 + 2*16) g/mol = 44 g/mol.
• molecular wt. of CH4 = (12 + 4*1) g/mol. = 16 g/mol.
• molecular wt. of O2 = 2*16 g/mol. = 32 g/mol.
• molecular wt. of H2O = (2*1 + 16) g/mol. = 18 g/mol.

data provided in the question is :-

• amount of CO2 obtained = 5 g
• amount of CH4 = 10 g
• amount of O2 = 10 g

formula used :-

yield % = (actual yield / theoretical yield) * 100

Reaction given:-

CH4 + 2O2 CO2 + 2H2O

from the reaction :-

2*32 g of O2 reacts with 16 g of CH4

10 g of O2 reacts with {(16/64)*10 } g of CH4

= (1/4)*10 g of CH4

= 2.5 g of CH4

again from reaction,

16 g of CH4 produce 44 g of CO2 so,

2.5 g of CH4 produce {(44/16) * 2.5 } g of CO2

=( 2.75 * 2.5 ) g of CO2

= 6.875 g of CO2

so,

theoretical yield of CO2 = 6.875 g

actual yield of CO2 = 5 g

so,

yield % =( actual yield of CO2/theoretical yield of CO2) * 100

= ( 5 g / 6.875 g) * 100

= 0.7272 * 100

= 72.72%

yield % = 72.72 %