How many grams of water are produced from the complete combustion of 44.0 grams of propane?

• A. 36.0 g
• B. 72.0 g
• C. 50.0 g
• D. 94.0 g

Answer: (B)

To determine how many grams of water are produced from the complete combustion of 44.0 grams of propane (C3H8), we first need to understand the combustion reaction.

The combustion of propane can be represented by the balanced chemical equation:

[

C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O

]

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

Next, we need to convert grams of propane into moles. The molar mass of propane is calculated as follows:

• Carbon: 12.01 g/mol × 3 = 36.03 g/mol

• Hydrogen: 1.008 g/mol × 8 = 8.064 g/mol

Adding these together gives:

[

36.03 + 8.064 = 44.094 , \text{g/mol}

]

(For practical purposes, we often round this to approximately 44.0 g/mol).

Now, using the given mass of propane (44.0 grams), we find the number of moles of propane:

[

\text{Moles of propane} = \frac