C5H12+802->5CO2+6H2O

Question

how many grams of water can be produced from 15 grams of C5H12 and 20 grams of Oxygen? What is the limiting reactant?

Answer 1

With 15g of C₅H₁₂ you would predict (15g C₅H₁₂)*(1 mol C₅H₁₂ /72.1g C₅H₁₂)*(6 mol H₂O /1 mol C₅H₁₂) = 1.25 moles of H₂O.

With 20g of O₂ gas, you would predict (20g O₂)*(1 mol O₂ /32.0g O₂)*(6 mol H₂O /8 mol O₂) = 0.469 mole of H₂O. This is limiting, because when you have made this much water, there is no more oxygen to consume.

Finally, (0.469 mole H₂O)*(18.0g H₂O/1 mol H₂O) = 8.4g H₂O (to 2 sig figs).