Question

Help please. I'm not good at Chem. :(

C3H8 + 5O2 --> 4H2O + 3CO2

If 0.45 grams of water is formed during the combustion of propane, how many grams of carbon dioxide is formed?

Answer 1

There will be 0.825 grams of CO2 produced

Step-by-step explanation:

Step 1: Data given

Mass of water formed = 0.45 grams

Molar mass of water = 18.02 g/mol

Molar mass of CO2 = 44.01 g/mol

Step 2: The balanced equation

C3H8 + 5O2 → 4H2O + 3CO2

Step 3: Calculate moles of water

Number of moles water = mass of water / molar mass water

Number of moles water = 0.45 grams / 18.02 g/mol

Number of moles water = 0.025 moles

Step 4: Calculate moles of CO2

For 1 mol of propane we need 5 moles of O2 to produce 4 moles of H2O and 3 moles of CO2

When 0.025 moles of H2O are produced, we'll get 3/4 * 0.025 = 0.01875 moles of CO2 produced

Step 5: Calculate mass of CO2 produced

Mass of CO2 = moles of CO2 * molar mass of CO2

Mass of CO2 = 0.01875 moles * 44.01 g/mol

Mass of CO2 = 0.825 grams

There will be 0.825 grams of CO2 produced