Answer:
A. 70.7%
Explanation:
In the first step lets compute the molar mass of CH₃OH and CO
Molar Mass of CH₃OH = 1(12.01 g/mol) + 4(1.008 g/mol) +1(16.00 g/mol)
= 32.042 g/mol
Molar Mass of CO₂ = 1(12.01 g/mol) + 2(16.00 g/mol)
= 44.01 g/mol
Mass of only one reactant i.e. CH₃OH is given so it must be the limiting reactant. Next, the theoretical yield is calculated directly as follows:
Given mass of CH₃OH is 10.4 g. So we have:
10.4g CH₃OH
Convert grams of CH₃OH to moles of CH₃OH utilizing molar mass of CH₃OH as:
1 mol CH₃OH / 32.042 g CH₃OH
Convert CH₃OH to moles of CO₂ using mole ratio as:
2 mol CO₂ / 2 mol CH₃OH
Convert moles of CO₂ to grams of CO₂ utilizing molar mass of CO₂ as:
44.01 g/mol CO₂ / 1 mol CO₂
Now calculating theoretical yield using above steps:
[ 10.4 g CH₃OH ] [1 mol CH₃OH / 32.042 g CH₃OH ] [2 mol CO₂ / 2 mol CH₃OH] [44.01 g/mol CO₂ / 1 mol CO₂]
Multiplication is performed here. We are left with 10.4 and 44.01 g CO₂ from numerator terms in the above equation and 32.042 from denominator terms after cancellation process of above terms. So this equation becomes:
= ( 10.4 ) ( 44.01 ) g CO₂ / 32.042
= 457.704/32/042
= 14.28 g CO₂
Theoretical yield = 14.28 g CO₂
Finally compute the percent yield for a process in which 10.4g of CH₃OH reacts and 10.1 g of CO₂ is formed:
percent yield = (actual yield / theoretical yield) x 100
As we have calculated theoretical yield which is 14.28 g CO₂ and actual yield is 10.1 g CO₂ So,
percent yield = (10.1 g CO₂ / 14.28 g CO₂) x 100%
= 0.707 x 100%
= 70.7 %
Hence option A 70.7% yield is the correct answer.
