Question

Reaction of \(2.0 \mathrm{~L}\) of hydrogen gas with \(1.0 \mathrm{~L}\) of oxygen gas yields \(2.0 \mathrm{~L}\) of water vapor. All gases are at the same temperature and pressure. Show how these data support the idea that oxygen gas is a diatomic molecule. Must we consider hydrogen to be a diatomic molecule to explain these results?

Short answer

The balanced chemical equation for this reaction is \(2 H_2 (g) + O_2 (g) \rightarrow 2 H_2 O (g)\). From the given volume ratios and applying Avogadro's Law, we can deduce that the ratios follow the stoichiometric coefficients in the balanced equation, which supports the idea that oxygen gas is a diatomic molecule. The balanced equation also indicates that hydrogen gas, being present as a diatomic molecule with a stoichiometric coefficient of 2, reacts with oxygen gas accordingly. Thus, we must consider hydrogen to be a diatomic molecule to explain these results.

Step-by-step solution

Write down the balanced chemical equation for the given reaction

The balanced chemical equation for the given reaction is: \(2 H_2 (g) + O_2 (g) \rightarrow 2 H_2 O (g)\) The coefficients of the balanced equation represent the mole ratios in which the reactants combine to form products.

Analyze the volume ratios

According to Avogadro's Law, equal volumes of gases at the same temperature and pressure contain the same number of molecules or moles. Since the volume of hydrogen gas is \(2.0 \mathrm{~L}\), the volume of oxygen gas is \(1.0 \mathrm{~L}\), and the volume of water vapor is \(2.0 \mathrm{~L}\), we can deduce the following ratios: \(2.0 \mathrm{~L} \frac{H_{2}}{1.0 \mathrm{~L} O_{2}} = 2.0 \mathrm{~L} \frac{H_{2}O}{1.0 \mathrm{~L} O_{2}}\) These volume ratios are equal to the mole ratios due to Avogadro's Law. This indicates that for every \(2\) moles of hydrogen gas that react, \(1\) mole of oxygen gas is consumed, and \(2\) moles of water vapor are produced.

Relate the mole ratios to the stoichiometric coefficients

From the balanced chemical equation, we can see that the stoichiometric coefficients are as follows: \(2 H_2: 1 O_2: 2 H_2O\) Comparing these coefficients to the volume ratios we calculated in step 2, we can see that the ratios are the same: \(2.0 \mathrm{~L} \frac{H_{2}}{1.0 \mathrm{~L} O_{2}} = 2.0 \mathrm{~L} \frac{H_{2}O}{1.0 \mathrm{~L} O_{2}}\) These equal ratios support the idea that oxygen gas is a diatomic molecule, as it combines with hydrogen gas in a 1:2 ratio, which also corresponds to the balanced chemical equation.

Determine if hydrogen is a diatomic molecule

From the balanced chemical equation, we see that hydrogen gas is also present as a diatomic molecule because the stoichiometric coefficient of hydrogen gas is 2. This means that 2 moles of hydrogen gas react with 1 mole of oxygen gas to produce 2 moles of water vapor, which is consistent with the volume ratios calculated in step 2: \(2.0 \mathrm{~L} \frac{H_{2}}{1.0 \mathrm{~L} O_{2}} = 2.0 \mathrm{~L} \frac{H_{2}O}{1.0 \mathrm{~L} O_{2}}\) This is further supported by the fact that hydrogen gas is commonly present as a diatomic molecule (\(H_{2}\)) due to its lower energy state and increased stability. Therefore, we must consider hydrogen to be a diatomic molecule to explain these results.