Question

Calculate the mass of water produced when 7.63 g of butane reacts with excess oxygen.

Answer 1

C₄H₁₀ +
`(13)/(2)`O₂ → 4CO₂ + 5H₂O

If mass of butane = 7.63 g
and mole = mass ÷ molar mass

then mole of butane = 7.63 g ÷ ((12 × 4) + (1 × 10))
= 7.63 g ÷ 58
= 0.1316 mol

Mole Ratio of C₄H₁₀ : H₂O is 1 : 5

∴ if mole of butane = 0.1316 mol

then mole of water = (0.1316 mol × 5 )
= 0.658 mol

Since mass = moles × molar mass

then mass of water produced = 0.658 mol × ((1 × 2) + (16 × 1))
= 11.844 g

Answer 2

You can calculate the mass of water by using the molar units to convert 7.63g of butane to moles and then find the molar mass of H2O which I think should be 1.01 + 1.01 + 16.00 if I remember correctly.  That is 18.02moles/g.  The butane is the limiting reactant but you would need the chemical equation to solve this