Question

Calculate the mass of water produced when 2.15 g of butane reacts with excess oxygen in the equation 2C4H10(g)+13O2(g)→10H2O(g)+8CO2(g)

Answer 1

molar mass :

H₂O = 18.0 g/mol

C₄H₁₀ = 58.0 g/mol

2 C₄H₁₀(g)+ 13 O₂(g) →10 H₂O(g)+ 8 CO₂(g)

2 x ( 58.0) g -------------- 10 x ( 18.0 ) g
2.15 g -------------------- mass of water

mass of water = 2.15 x 10 x 18.0 / 2 x 58.0

mass of water = 387 / 116

mass of water = 3.336 g

hope this helps!