Question

2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g) Calculate the mass of water produced when 4.16 g of butane reacts with excess oxygen.

Answer 1

Ans-- given balence reaction is

2C₄H₁₀(g)+13O₂(g)→10H₂O(g)+8CO₂(g)

we can use dimensional analysis, and equation coefficients, to change 4.16 g of butane to moles of butane, to moles of water, to grams of water. We shall need the following equalities to set up conversion factors.

1 mol C4H10 = 56.108 g C4H10
2 mol C4H10 = 10 mol H2O
1 mol H2O = 18.015 g H2O

[(4.16 g C4H10)/1][(1 mol C4H10)/(56.108 g C4H10)][(10 mol H2O)/(2 mol C4H10)][(18.015 g H2O)/(1 mol H2O)] = 6.38248 g of H2O or 6.39 g H2O rounded to three significant figures

Answer: The mass of water produced when 4.16 of butane reacts with excess oxygen is 6.39 g.