Question

Consider the combustion of ethane:

2 C2H6 + 7 O2
4 CO2 + 6 H2O

If 60 grams of ethane (molar mass of 30 grams/mol) react, how many moles of H2O are produced?

Group of answer choices

9

6

3

12

Answer 1

To determine how many moles of water (H2O) are produced from the combustion of ethane (C2H6), we can follow these steps:

1. **Calculate the moles of ethane (C2H6) reacted**:
- Given that the molar mass of C2H6 is 30 grams/mol, we can calculate the moles of ethane in 60 grams:
\[
\text{Moles of C2H6} = \frac{\text{mass of C2H6}}{\text{molar mass of C2H6}} = \frac{60 \text{ g}}{30 \text{ g/mol}} = 2 \text{ moles of C2H6}
\]

2. **Use the stoichiometry of the reaction**:
- The balanced equation for the combustion of ethane is:
\[
2 \text{ C2H6} + 7 \text{ O2} \rightarrow 4 \text{ CO2} + 6 \text{ H2O}
\]
- According to the equation, 2 moles of C2H6 produce 6 moles of H2O.

3. **Calculate the moles of H2O produced**:
- From the stoichiometry of the reaction:
\[
\text{Moles of H2O produced} = \left(\frac{6 \text{ moles of H2O}}{2 \text{ moles of C2H6}}\right) \times \text{moles of C2H6 reacted}
\]
- Since we have 2 moles of C2H6:
\[
\text{Moles of H2O produced} = \left(\frac{6}{2}\right) \times 2 = 6 \text{ moles of H2O}
\]

Therefore, if 60 grams of ethane react, **6 moles of H2O** are produced.

The correct answer is **6**.