33.6k views
1 vote
What is the theoretical yield of methanol (CH3OH) when 12.0 grams of H2 is mixed with 74.5 grams of CO? CO + 2H2 CH3OH

A) 14.88 grams
B) 47.65 grams
C) 74.50 grams
D) 85.12 grams
E) 92.04 grams

1 Answer

2 votes
1) We need to convert 12.0 g of H2 into moles of H2, and 74.5 grams of CO into moles of CO
Molar mass of H2: M(H2) = 2*1.0= 2.0 g/mol
Molar mass of CO: M(CO) = 12.0 +16.0 = 28.0 g/mol

12.0 g H2 * 1 mol/2.0 g = 6.0 mol H2
74.5 g CO * 1 mol/28.0 g = 2.66 mol CO

2) Now we can use reaction to find out what substance will react completely, and what will be leftover.

CO + 2H2 -------> CH3OH
1 mol 2 mol
given 2.66 mol 6 mol (excess)

How much
we need CO? 3 mol 6 mol

We see that H2 will be leftover, because for 6 moles H2 we need 3 moles CO, but we have only 2.66 mol CO.
So, CO will react completely, and we are going to use CO to find the mass of CH3OH.

3)
CO + 2H2 -------> CH3OH
1 mol 1 mol
2.66 mol 2.66 mol

4) We have 2.66 mol CH3OH
Molar mass CH3OH : M(CH3OH) = 12.0 + 4*1.0 + 16.0 = 32.0 g/mol

2.66 mol CH3OH * 32.0 g CH3OH/ 1 mol CH3OH = 85.12 g CH3OH

Answer is
D) 85.12 grams.
User SyntaxError
by
7.8k points