Question

the aluminum is allowed to react with the sulfuric reaction until it stops reacting, and there is still some unreacted aluminum in the reaction flask.
the gas in the reaction flask is brought to the same atmospheric pressure as the surroundings, 743.9 torr and 23.0°c. the vapor pressure of water at 23.0°c is 21.1 torr. the collected gas is found to have a volume of 652 ml.
what is the partial pressure of the hydrogen gas in the collection flask? torr
how many moles of hydrogen gas are contained in the collection flask? mol
what mass of aluminum reacted? g (hint: you need to determine the balanced equation for the reaction to do this problem.)

Explanation:

Step1: Calculate partial pressure of hydrogen gas

According to Dalton's law of partial pressures, \(P_{total}=P_{H_2}+P_{H_2O}\). So \(P_{H_2}=P_{total}-P_{H_2O}\).
Given \(P_{total} = 743.9\) torr and \(P_{H_2O}=21.1\) torr.
\(P_{H_2}=743.9 - 21.1=722.8\) torr.

Step2: Calculate moles of hydrogen gas

Use the ideal gas law \(PV = nRT\). First, convert units: \(V = 652\space mL=0.652\space L\), \(T=(23.0 + 273.15)K = 296.15\space K\), \(R = 62.36\space L\cdot torr/(K\cdot mol)\) (since pressure is in torr), \(P = 722.8\) torr.
Rearrange for \(n\): \(n=\frac{PV}{RT}\)
\(n=\frac{722.8\times0.652}{62.36\times296.15}\)
\(n=\frac{471.2656}{18465.374}\approx0.0255\space mol\)

Step3: Calculate mass of aluminum reacted

The balanced chemical equation for the reaction of \(Al\) with \(H_2SO_4\) is \(2Al + 3H_2SO_4=Al_2(SO_4)_3+3H_2\).
From the equation, the mole ratio of \(Al\) to \(H_2\) is \(2:3\). So moles of \(Al\) reacted, \(n_{Al}=\frac{2}{3}n_{H_2}\)
\(n_{Al}=\frac{2}{3}\times0.0255 = 0.017\space mol\)
Molar mass of \(Al\) is \(M = 26.98\space g/mol\). Mass \(m = n\times M\)
\(m=0.017\times26.98 = 0.459\space g\)

Answer:

• Partial pressure of hydrogen gas: \(722.8\) torr
• Moles of hydrogen gas: \(0.0255\) mol
• Mass of aluminum reacted: \(0.459\) g