In the complete combustion of 44.0 grams of propane, how many grams of water are formed?

In the complete combustion of 44.0 grams of propane, how many grams of water are formed?

• A. 4.50g
• B. 18.0g
• C. 72.0g
• D. 176g

Answer: (C)

To determine how many grams of water are formed from the complete combustion of propane (C3H8), we need to first examine the balanced chemical equation for this reaction. The combustion of propane can be represented as follows: \[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} \] From the balanced equation, we can see that one mole of propane produces four moles of water. Next, we need to calculate the moles of propane in 44.0 grams. The molar mass of propane (C3H8) is approximately 44.1 g/mol (calculating from the atomic masses of carbon and hydrogen, i.e., 12.01 g/mol for carbon and 1.008 g/mol for hydrogen). Thus, for 44.0 grams of propane: \[ \text{Moles of propane} = \frac{44.0 \text{ g}}{44.1 \text{ g/mol}} \approx 1.0 \text{ mol} \] According to the stoichiometry from the balanced equation,

To determine how many grams of water are formed from the complete combustion of propane (C3H8), we need to first examine the balanced chemical equation for this reaction. The combustion of propane can be represented as follows:

[ \text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O} ]

From the balanced equation, we can see that one mole of propane produces four moles of water.

Next, we need to calculate the moles of propane in 44.0 grams. The molar mass of propane (C3H8) is approximately 44.1 g/mol (calculating from the atomic masses of carbon and hydrogen, i.e., 12.01 g/mol for carbon and 1.008 g/mol for hydrogen). Thus, for 44.0 grams of propane:

[ \text{Moles of propane} = \frac{44.0 \text{ g}}{44.1 \text{ g/mol}} \approx 1.0 \text{ mol} ]

According to the stoichiometry from the balanced equation,