Calculate the mass of water produced when 3.03 g of butane reacts with excess oxygen.

The mass of water produced is 4.70 g.  

M_r:   58.12     32.00  

         2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O

m/g:  3.03

Moles of C₄H₁₀ = 3.03 g C₄H₁₀ × (1 mol C₄H₁₀/58.12 g C₄H₁₀) = 0.05 213 mol C₄H₁₀

Moles of H₂O = 0.052 13 mol C₄H₁₀ × (10 mol H₂O/2 mol C₄H₁₀) = 0.2607 mol H₂O

Mass of H₂O = 0.2607 mol H₂O × (18.02 g H₂O/1 mol H₂O) = 4.70 g H₂O

first one is the correct answer