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Elina [12.6K]
3 years ago
8

Calculate the mass of water produced when 6.97 g of butane reacts with excess oxygen

Chemistry
1 answer:
andrew-mc [135]3 years ago
7 0
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O 
the limiting reactant in this reaction is C₄H₁₀  This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
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