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

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