Question

A 14.22-g aluminum soda can reacts with hydrochloric acid according to the reaction:

2Al(s)+6HCl(aq)3H2(g)+2AlCl3(aq)2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(a q) \longrightarrow 3 \mathrm{H}_2(g)+2 \mathrm{AlCl}_3(a q)

The hydrogen gas is collected by the displacement of water. What volume of hydrogen is collected if the external pressure is 749 mmHg749 \mathrm{~mm} \mathrm{Hg} and the temperature is 31C31{ }^{\circ} \mathrm{C}, assuming that there is no water vapor present? What is the vapor pressure of water at this temperature? What volume do the hydrogen and water vapor occupy under these conditions?

Solution

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The given balanced chemical reaction is shown below:

2Al(s) + 6HCl(aq)2AlCl3(aq) + 3H2(g)\begin{aligned}\text{2Al}_{\text{(s)}}~+~\text{6HCl}_{\text{(aq)}}\rightarrow\text{2AlCl}_{3\text{(aq)}}~+~\text{3H}_{2\text{(g)}}\end{aligned}

One of the products of the reaction is H2\text{H}_{2} gas. When the gas product of a chemical reaction is collected over water, also known as the water displacement method, the collected gas would be a mixture of water and the gas product. This is because some water molecules will evaporate, and the amount of the water molecules that will evaporate will depend on the temperature of the reaction. Since the collected gas is a gas mixture, the total pressure of the gas mixture (Ptotal)(P_{total}) would be equal to the partial pressure of the hydrogen gas (PH2P_{\text{H}_{2}}) and the partial pressure of water (PH2OP_{\text{H}_{2}\text{O}}), which is also called as the vapor pressure. Therefore:

Ptotal=PH2 + PH2O\begin{aligned}P_{total} = P_{\text{H}_{2}}~+~P_{\text{H}_{2}\text{O}}\end{aligned}

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