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) 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.