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

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

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JoshEast JoshEast · Answer 2 · 2017-03-16T21:00:33-04:00

C₄H₁₀ + \(\frac{13}{2}\)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