If 41.4 grams of propane are combusted with 32.2 grams of oxygen, how many grams of water will be formed?

Question

C₃H₈ (g) + 5 O₂ (g) ------> 3 CO₂ (g) + 4 H₂O (g)

Balanced Equation

Answer 1

C₃H₈ (g) + 5 O₂ (g) ------> 3 CO₂ (g) + 4 H₂O (g)

moles propane = 41.4 g x 1 mole/44g = 0.941 moles

moles oxygen = 32.2 g x 1 mole/32 g = 1 mole

According to the balanced equation it takes 5 moles O₂ for each mole of propane. Thus, oxygen is limiting.

g H₂O = 1 mol O₂ x 4 mol H₂O/mol O₂ x 18 g/mol = 72 g

Note: I rounded off when using molar masses but you shouldn’t. I reported the answer to 2 sig figs, but you should use 3 sig figs.