The synthesis of methanol from hydrogen and carbon monoxide takes place following the chemical reaction
2*H2 +CO = CH3OH
The molecular masses of the reactants and of the reaction product are
M(H2) =2 g
M(CO) =12 +16 =28 g
M(CH3OH) = 12+16 +4 = 32 g
Thus for 65.8 grams of CO corresponds 65.8/28 = 2.35 moles of CO
For 8.6 grams of H2 corresponds 8.6/2 = 4.3 moles of H2
Since in the reaction above for 1 mole of CO entering the reaction we need 2 moles of H2 as a by-reactant, for 2.35 moles of CO (65.8 grams) one needs 2*2.35 =4.7 moles of H2 to enter reaction (and we have only 4.3 moles H2). Therefore hydrogen is the limiting reactant here.
From the above reaction we can write the following mass balance
2*2 grams of H2 gives......32 grams CH3OH
8.6 grams of H2 gives ......x grams CH3OH
x =32*8.6/4 =68.8 grams CH3OH
The number of moles of CH3OH in 68.8 grams of substance is
n =68.8/32 =2.15 moles CH3OH
The theoretical yield of the reaction is 68.8 grams (2.15 moles) of CH3OH.