Limiting Reactant, Production, and Excess Reactant

Question

If 16.8g of carbon monoxide (CO) is mixed under high pressure with 3.78g of hydrogen gas (H2) to produce methanol, what would be the limiting reactant? What mass of Methanol will be produced? How much Excess reactant would remain?

Answer 1

1) Write the balanced chemical reaction.

CO + 2 H2 → CH3OH

2) Use stoichiometry to convert g of each reactant to g of methanol. The reactant that produces the smaller amount of methanol is the limiting reagent.

16.8 g CO x (1 mol CO / 28 g CO) x (1 mol CH3OH / 1 mol CO) x (32 g CH3OH / 1 mol CH3OH) = 19.2 g CH3OH

3.78 g H2 x (1 mol H2 / 2 g H2) x (1 mol CH3OH / 2 mol H2) x (32 g CH3OH / 1 mol CH3OH) = 30.2 g CH3OH

CO produces the lesser amount of CH3OH, therefore, it is the limiting reagent.

3) From above, 19.2 g of CH3OH will be produced from this reaction.

4) Convert g of CO to g of H2 to see how much will react.

16.8 g CO x (1 mol CO / 28 g CO) x (2 mol H2 / 1 mol CO) x (2 g H2 / 1 mol H2) = 2.40 g H2

Out of 3.78 g of H2 that you start with, only 2.40 g will react with 16.8 g of CO.

Excess H2 = 3.78 - 2.40 = 1.38 g H2 remaining