Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.

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asked May 5, 2018 51.7k views

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Doug Glancy · Answer 1 · 2018-05-11T03:18:01+0000

c4h10+6.5o2=4co2+5h2o
moles of butane=1.92/58=0.0331 moles
moles of water=0.1655 moles\
as the butane and water has 1 is to 5 molar ratio
0.1655=mass/18
mass=2.98 g
mass of water produced = 2.98 g

Calculate the mass of water produced when 1.92 g of butane reacts with excess oxygen.

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