How do you solve this question?

Question

Methanol is used as a fuel in racecars. It can be manufactured according to the reaction (10 points): _____H2 (g) + _____CO (g) _____ CH3OH (l)

If 68.3 g of CO is reacted with 8.60 g of H2, how much methanol will be formed?

Answer 1

_____H2 (g) + _____CO (g) _____ CH3OH (l)

So, first we have to balance this reaction. We do this by putting a 2 in front of the H2(g). So, we have...

2H₂(g) + CO(g) ==> CH₃OH(l) ... balanced equation

Next, we find which reactant is in limiting supply.

mols H₂ present = 8.60 g H₂ x 1 mol H2 / 2 g = 4.30 mols H₂

mols CO present = 68.3 g CO x 1 mol CO/ 28 g = 2.44 mols CO

Since according to the balanced equation it takes twice as much H₂ as CO, H₂ will be limiting as there isn't twice as much and it will run out first.

So now we use the mols of the limiting reactant (H₂) to find how much CH₃OH can be made.

4.30 mols H₂ x 1 mol CH₃OH / 2 mols H₂ = 2.15 mols CH₃OH x 32.0 g / mol = 68.8 g CH₃OH formed