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How many grams of CO2 and H2O are produced from the combustion of 220. g of propane (C3H8)? C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

User Kimmon
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Answer:

(C3H8) produces 660 g of CO2 and 360 g of H2O

Step-by-step explanation:

The balanced chemical equation for the combustion of propane (C3H8) is:

C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

This equation tells us that for every molecule of propane (C3H8) that reacts with 5 molecules of oxygen (O2), 3 molecules of carbon dioxide (CO2) and 4 molecules of water (H2O) are produced.

So, if we have 220. g of propane (C3H8), we can find the amount of CO2 and H2O produced by using the mole ratio from the balanced equation:

1 mole C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O

We can find the number of moles of C3H8 by dividing the mass by the molar mass of C3H8 (44 g/mol):

220 g / 44 g/mol = 5 moles C3H8

So, the number of moles of CO2 and H2O produced can be found by multiplying the number of moles of C3H8 by the mole ratio:

3 moles CO2 = 3 moles CO2/1 mole C3H8 * 5 moles C3H8 = 15 moles CO2

4 moles H2O = 4 moles H2O/1 mole C3H8 * 5 moles C3H8 = 20 moles H2O

Finally, we can convert the number of moles of CO2 and H2O to grams by multiplying by their molar masses (44 g/mol for CO2 and 18 g/mol for H2O):

15 moles CO2 * 44 g/mol = 660 g CO2

20 moles H2O * 18 g/mol = 360 g H2O

So, the combustion of 220 g of propane (C3H8) produces 660 g of CO2 and 360 g of H2O.

User Jesse Black
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