Aluminum metal (2.00g) react rapidly with aqueous H2SO4 acid to produce aqueous aluminum sulphate and hydrogen gas

Question

Calculate the mass of aluminium sulphate produced
Calculate the numbers of moles of H2SO4 used
Determine the volume of H2

Answer 1

To solve the problem, we need to start by writing the balanced chemical equation for the reaction between aluminum metal (Al) and sulfuric acid (H₂SO₄) to form aluminum sulfate (Al₂(SO₄)₃) and hydrogen gas (H₂).

1. Balanced Chemical Equation: \[ 2 \text{Al} + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2 \]

2. Calculate the number of moles of aluminum (Al): First, we need to calculate the number of moles of Al using its molar mass. The molar mass of Al is approximately 26.98 g/mol.

   \[ \text{Moles of Al} = \frac{\text{mass of Al}}{\text{molar mass of Al}} = \frac{2.00 , \text{g}}{26.98 , \text{g/mol}} \approx 0.0741 , \text{mol} \]

3. Calculate the moles of H₂SO₄ used: From the balanced equation, we see that 2 moles of Al react with 3 moles of H₂SO₄. To find the moles of H₂SO₄ required for 0.0741 moles of Al:

   \[ \text{Moles of H}_2\text{SO}_4 = \left(\frac{3 \text{ moles H}_2\text{SO}_4}{2 \text{ moles Al}}\right) \times 0.0741 , \text{mol Al} \approx 0.1112 , \text{mol H}_2\text{SO}_4 \]

4. Calculate the mass of aluminum sulfate produced: From the balanced equation, 2 moles of Al produce 1 mole of Al₂(SO₄)₃. Therefore, 0.0741 moles of Al will produce:

   \[ \text{Moles of Al}_2(\text{SO}_4)_3 = \left(\frac{1 \text{ mole Al}_2(\text{SO}_4)_3}{2 \text{ moles Al}}\right) \times 0.0741 , \text{mol Al} = 0.03705 , \text{mol Al}_2(\text{SO}_4)_3 \]

   Next, we calculate the molar mass of aluminum sulfate (Al₂(SO₄)₃): \[ \text{Molar mass} = (2 \times 26.98) + (3 \times (32.07 + 4 \times 16.00)) = 53.96 + 3 \times (32.07 + 64.00) = 53.96 + 3 \times 96.07 = 53.96 + 288.21 = 342.17 , \text{g/mol} \]

   Now calculate the mass of aluminum sulfate produced: \[ \text{Mass of Al}_2(\text{SO}_4)_3 = \text{moles} \times \text{molar mass} = 0.03705 , \text{mol} \times 342.17 , \text{g/mol} \approx 12.68 , \text{g} \]

5. Determine the volume of H₂ produced: From the balanced reaction, 2 moles of Al produce 3 moles of H₂. Therefore, moles of H₂ produced from 0.0741 moles of Al will be:

   \[ \text{Moles of H}_2 = \left(\frac{3 \text{ moles H}_2}{2 \text{ moles Al}}\right) \times 0.0741 , \text{mol Al} \approx 0.1112 , \text{mol H}_2 \]

   At standard temperature and pressure (STP), 1 mole of gas occupies 22.4 liters. Therefore, the volume of H₂ produced:

   \[ \text{Volume of H}_2 = \text{moles} \times 22.4 , \text{L/mol} \approx 0.1112 , \text{mol} \times 22.4 , \text{L/mol} \approx 2.49 , \text{L} \]

Summary:

1. Mass of aluminum sulfate produced: \(\approx 12.68 , \text{g}\)
2. Number of moles of H₂SO₄ used: \(\approx 0.1112 , \text{mol}\)
3. Volume of H₂ produced: \(\approx 2.49 , \text{L}\)