Question

a student performs an experiment to determine the volume of hydrogen gas produced when a given mass of magnesium reacts with excess hcl(aq), as represented by the net ionic equation above. the student begins with a 0.0360 g sample of pure magnesium and a solution of 2.0 m hcl(aq).
a. calculate the number of moles of magnesium in the 0.0360 g sample.
b. calculate the number of moles of hcl(aq) needed to react completely with the sample of magnesium.
c. as the magnesium reacts, the hydrogen gas produced is collected by water displacement at 23.0°c. the pressure of the gas in the collection tube is measured to be 749 torr. given that the equilibrium vapor pressure of water is 21 torr at 23.0°c, calculate the pressure that the h2(g) produced in the reaction would have if it were dry.

Explanation:

Step1: Calculate moles of magnesium

The molar mass of \(Mg\) is \(M = 24.305\space g/mol\).
Using the formula \(n=\frac{m}{M}\), where \(m = 0.0360\space g\) and \(M = 24.305\space g/mol\).
\(n_{Mg}=\frac{0.0360\space g}{24.305\space g/mol}\approx1.48\times 10^{-3}\space mol\)

Step2: Calculate moles of \(HCl\)

The balanced chemical equation is \(Mg + 2HCl=MgCl_{2}+H_{2}\).
From the stoichiometry, the mole ratio \(n_{HCl}:n_{Mg}=2:1\).
Since \(n_{Mg}\approx1.48\times 10^{-3}\space mol\), then \(n_{HCl}=2\times n_{Mg}\).
\(n_{HCl}=2\times1.48\times 10^{-3}\space mol = 2.96\times 10^{-3}\space mol\)

Step3: Calculate pressure of dry \(H_{2}\)

The total pressure \(P_{total}=749\space torr\) and the vapor - pressure of water \(P_{H_{2}O}=21\space torr\).
Using Dalton's law \(P_{H_{2}}=P_{total}-P_{H_{2}O}\).
\(P_{H_{2}}=(749 - 21)\space torr=728\space torr\)

Answer:

a. \(n_{Mg}\approx1.48\times 10^{-3}\space mol\)
b. \(n_{HCl}\approx2.96\times 10^{-3}\space mol\)
c. \(P_{H_{2}} = 728\space torr\)