Question

1.53 g of aluminum metal is placed in a Erlenmeyer flask with 1.5 M sulfuric acid. The hydrogen gas produced from the reaction is collected over water as seen in the diagram below.

The aluminum is allowed to react with the sulfuric acid until it stops reacting. There is still some unreacted aluminum left in the reaction flask. The gas in the collection flask is brought to the same atmospheric pressure as the surroundings, 765.25 torr and a temperature of 24.0 ^oC. The vapor pressure of water at 24.0 ^oC is 22.4 torr. The collected gas is found to have a volume of 350 mL.

Answer the following three questions based on this scenario.

What is the partial pressure of the hydrogen in the collection flask?

1. 742.8 torr
2. 787.6 torr
3. 765.25 torr
4. 760 torr

How many moles of hydrogen gas are contained in the collection flask?

1. 14 mol
2. 0.024 mol
3. 0.014 mol
4. 0.17 mol
5. 11 mol
6. 174 mol

What mass of aluminum reacted? (hint: You need to determine the balanced equation for the reaction to do this problem.)

? g Al

Answer 1

$\large \bf (1)$

The gas that is collected is a mixture of H, and H2O vapor. Determine pressure of the H, (only, Ph.) by subtracting the vapor pressure of water (PH.0) from P Total

PH, PTotal Рн.0

PTotal 765.25 torr, Ph,o = 22.4 torr

Рн, 765.25 torr – 22.4 torr

$\large \bf P_{H_2}=742.8\;torr$

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$\large \bf (2)$

$\text{Using the ideal gas equation PV=nRT}$

$\bf P_{H_2}=742.8\;torr=0.977\;atm, \;V_{H_2O}=350\;mL=0.350\;L,\;T=24^oC=297.15K$

$\bf n_{H_2}=\frac{P_{H_2}V_{H_2}}{RT}$

$\bf n_{H_2}=\frac{0.977\;atm\times0.350\;L}{0.0821\;L.atm/mol.K\times297.15\;K}$

$\large \bf n_{H_2}=0.014 \; mol$

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$\large \bf (3)$

$\text{The balance equation is}$

$\bf 2Al(s)+3H_2SO_4(aq)\rightarrow Al_2(SO_4)_3(aq)+3H_2(g)$

$\text{3 moles of H}_2\text{ are produced from 2 moles of Al}$

$\text{0.014 moles of H}_2\text{ are produced from=} \frac{2}{3}\times0.014 \text{ moles of Al}$

$\text{0.014 moles of H}_2\text{ are produced from=} 0.0093 \text{ moles of Al}$

$\text{mass of Al reacted =0.0093 \cancel{ mol}}\times 27\text{ g/\cancel{mol}}$

$\large \textbf{mass of Al reacted = 0.251 g}$