Questions: Reference the following balanced equation: 2 H2(g) O2(g) ... 2 H2O(g) If you have 30.0 g of hydrogen gas burned in excess oxygen gas, how many moles of water can you theoretically make? 60 moles water 7.5 moles water 15 moles water 30 moles water

Reference the following balanced equation:
2 H2(g) O2(g) ... 2 H2O(g)

If you have 30.0 g of hydrogen gas burned in excess oxygen gas, how many moles of water can you theoretically make?
60 moles water
7.5 moles water
15 moles water
30 moles water

Transcript text: Reference the following balanced equation: \[ 2 \mathrm{H}_{2}(g) \quad \mathrm{O}_{2(g)} \quad \cdots \quad 2 \mathrm{H}_{2} \mathrm{O}_{(g)} \] If you have 30.0 g of hydrogen gas burned in excess oxygen gas, how many moles of water can you theoretically make? 60 moles water 7.5 moles water 15 moles water 30 moles water

Solution

Solution Steps

Step 1: Determine the Molar Mass of Hydrogen Gas

The molar mass of hydrogen gas (\(\mathrm{H}_2\)) is calculated as follows: \[ \text{Molar mass of } \mathrm{H}_2 = 2 \times 1.008 \, \text{g/mol} = 2.016 \, \text{g/mol} \]

Step 2: Calculate Moles of Hydrogen Gas

Using the given mass of hydrogen gas, calculate the number of moles: \[ \text{Moles of } \mathrm{H}_2 = \frac{30.0 \, \text{g}}{2.016 \, \text{g/mol}} = 14.88 \, \text{moles} \]

Step 3: Use Stoichiometry to Find Moles of Water

According to the balanced chemical equation: \[ 2 \mathrm{H}_2(g) + \mathrm{O}_2(g) \rightarrow 2 \mathrm{H}_2\mathrm{O}(g) \] The stoichiometric ratio of \(\mathrm{H}_2\) to \(\mathrm{H}_2\mathrm{O}\) is 1:1. Therefore, the moles of water produced will be equal to the moles of hydrogen gas used: \[ \text{Moles of } \mathrm{H}_2\mathrm{O} = 14.88 \, \text{moles} \]

Step 4: Determine the Closest Answer

The closest answer to 14.88 moles of water is 15 moles of water.