Can anyone answer these Stoichiometry quetions with work?

Question

Methyl alcohol (wood alcohol), CH3OH, is produced via the reaction
CO + 2H2 -> CH3OH
A mixture of 1.20 g H2 and 7.45 g CO are allowed to react.

(a) What reagent is the limiting reagent?

(b) What is the gram yield of CH3OH?

(c) How much of the reagent present in excess is left over?

(d) Suppose the actual yield is 7.52g of CH3OH. What is the % yield?

Answer 1

a) n(H2) = 1.20 / 2.00 = 0.600 mol.
n(CO) = 7.45 / 28.00 = 0.266 mol.
0.266 mol of CO requires 2×0.266 = 0.532 mol of H2 to react completely; there is more H2 than this, so CO is the limiting reagent and H2 is in excess.

b) n(CH3OH) = n(CO) = 0.266 mol
m(CH3OH) = 0.266(32.00) = 8.51 g.

c) m(H2) remaining = m(reactants) - m(products) = 8.65 - 8.51 = 0.14g. (Alternatively, we could say n(H2) remaining is 0.600 - 0.532 = 0.068 mol, and then m(H2) = 2.00(0.068) = 0.14g.)

d) 7.52 / 8.51 = 88.3%.

Answer 2

CO ==> 12 + 16 = 28 g/mol
H2 ==> 2*1 = 2 g/mol
CH3OH ==> 12 + 16 + 4*1 = 32 g/mol
Moles available:
CO = 7.45/28 = 0.26607142857142857142857142857143
H2 = 1.2/2 = 0.6
You need 2 M H2 per M CO, so divide 0.6 by 2 = 0.3
(a) The limiting reagent is CO
(b) x/32 = 7.45/28
x = 8.514286 g
(c)1.2 - 4*7.45/28 = 135.7 mg
(d) 7.52*100/8.514286 = 88.32%