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How many grams of H_{2}*O can be produced from the reaction of 100. g of C_{3}*H_{8}

User Rayan Sp
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1 Answer

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12 votes

Answer: 163.6g H2O

Step-by-step explanation:

The balanced chemical equation for the combustion of propane is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

This means 1 molecule of C3H8 react with 5 molecules of O2 to produce 3 molecules of CO2 and 4 molecules of H2O.

Molar mass C3H8 = 44 g/mol
According to the stochiometric coefficients 1 mol C3H8 will form 4H2O

100g C3H8 to mol = 100 g / 44 g/mol = 2.273 mol
This will produce 2.273*4 H2O = 9.09 mol H2O

Molar mass H2O = 18 g/mol
Mass of 9.09 mol H2O = 9.09 mol * 18 g/mol = 163.6 g H2O

Therefore, 163.6g of H2O will be formed from the combustion of 100g of C3H8

User Gokce
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