Answer:
a. The chemical equation for the reaction is:
2Al + 3H₂SO₄ → Al₂(SO₄)₃ + 3H₂
b. To find the mass of required sulfuric acid, we need to use stoichiometry. We can start by finding the number of moles of aluminum used in the reaction:
Molar mass of Al = 27 g/mol
Number of moles of Al = 5.4 g / 27 g/mol = 0.2 mol
According to the balanced equation, 3 moles of H₂SO₄ are required to react with 2 moles of Al. Therefore, the number of moles of H₂SO₄ required is:
Number of moles of H₂SO₄ = 3/2 x 0.2 mol = 0.3 mol
Molar mass of H₂SO₄ = 2 x 1 g/mol + 32 g/mol + 4 x 16 g/mol = 98 g/mol
Mass of H₂SO₄ required = 0.3 mol x 98 g/mol = 29.4 g
Therefore, 29.4 g of sulfuric acid is required to react with 5.4 g of aluminum.
c. To find the volume of hydrogen gas obtained, we need to use the ideal gas law:
PV = nRT
where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, R is the universal gas constant (0.0821 L atm/mol K), and T is the temperature in Kelvin.
We can start by finding the number of moles of hydrogen gas produced in the reaction. According to the balanced equation, 3 moles of H₂ are produced for every 2 moles of Al. Therefore, the number of moles of H₂ produced is:
Number of moles of H₂ = 3/2 x 0.2 mol = 0.3 mol
Assuming the reaction occurs at standard temperature and pressure (STP), which is 0°C (273 K) and 1 atm, we can use the molar volume of a gas at STP, which is 22.4 L/mol. Therefore:
V = nRT/P = 0.3 mol x 0.0821 L atm/mol K x 273 K / 1 atm = 6.58 L
Therefore, the volume of hydrogen gas produced at STP is 6.58 L.
Step-by-step explanation: