How many grams of water will form if 10.54g H2 react with 95.10 g O2

2H₂ + O₂ = 2H₂O

m (H₂) = 10.54 g

M (H₂) = 2.016 g/mol

m (O₂) = 95.10 g

M (O₂) = 31.999 g/mol

M (H₂O) = 18.015 g/mol

n (H₂) = m (H₂) / M (H₂)

n (H₂) = 10.54/2.016=5.228 mol

n (O₂) = m (O₂) / M (O₂)

n (O₂) = 95.10/31.999=2.972 mol

H₂:O₂=2:1

5.288:2.972 ⇒ 2:1.124 oxygen in excess

n (H₂O) = n (H₂)

m (H₂O) = n (H₂) M (H₂O)

m (H₂O) = 5.228*18.015=94.182 g