(i) Well, the limiting factor in this reaction is like the VIP of the party. It's the species that gets used up first and puts a stop to the party. In this case, we need to compare the number of moles of CO and H2 that we have.
To do that, we need to convert grams into moles using the molar mass. So, for CO, the molar mass is 28.01 g/mol. Dividing 356 g of CO by the molar mass gives us 12.71 mol.
For H2, the molar mass is 2.02 g/mol. Dividing 65 g of H2 by the molar mass gives us 32.18 mol.
Since the balanced equation is telling us that it takes 1 mol of CO to react with 2 mol of H2, we can see that we have more than enough H2 to react with the CO. Therefore, the limiting factor is CO.
(ii) Now, let's calculate the theoretical amount of methanol that could be produced. According to the balanced equation, 1 mol of CO reacts with 2 mol of H2 to produce 1 mol of CH3OH.
Since we have 12.71 mol of CO, we will end up with 12.71 mol of CH3OH.
To convert this into grams, we use the molar mass of methanol, which is 32.04 g/mol. Multiplying the mol by the molar mass, we get 407.43 g of methanol.
Therefore, the theoretical amount of methanol that could be produced is 407.43 g.
(iii) Lastly, let's calculate the percentage yield using the actual amounts. We were told that 352 g of methanol was produced.
The percentage yield is calculated using the formula: (actual yield / theoretical yield) x 100.
Plugging in the values, we get (352 g / 407.43 g) x 100 = 86.4%.
So, the percentage yield for the reaction is 86.4%. Better than average, but hey, this reaction could still use some improvement!