What is the total number of grams of h2o produced when 22 grams of propane undergoes complete combustion?

1 answer:

5 0

The combustion reaction of propane would be expressed as:

C3H8 + 5O2 = 3CO2 + 4H2O

To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:

moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O

Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.

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= 1000 x 10³ g / 226 g mol⁻¹
= 4424.78 mol

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= 6637.17 mol

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Hence, mass of Na₂CO₃ = 6637.17 mol x 106 g mol⁻¹
= 703540.02 g
= 703.540 kg

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Percentage yield = (650 kg / 703.540 kg) x 100%
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