Final answer:
The limiting reactant is magnesium (Mg) because it will be completely consumed first. The mass of excess reactant, sulfuric acid (H2SO4), is 9.72 g. The moles of magnesium sulfate (MgSO4) produced is 0.099 mol, and the mass of water produced is 3.561 g.
Step-by-step explanation:
To determine the limiting reactant, we need to compare the moles of each reactant to the balanced chemical equation. The balanced equation for the reaction is:
Mg + H2SO4 -> MgSO4 + H2
From the equation, we can see that 1 mole of Mg reacts with 1 mole of H2SO4 to produce 1 mole of MgSO4 and 1 mole of H2. We are given that 2.4 g of Mg and 0.2 mole of H2SO4 are used in the reaction.
To calculate the moles of Mg, we can use the molar mass of Mg (24.31 g/mol):
Moles of Mg = 2.4 g / 24.31 g/mol = 0.099 mol
Since the mole ratio between Mg and H2SO4 is 1:1, and we have 0.099 mol of Mg and 0.2 mol of H2SO4, we can see that Mg is the limiting reactant because it is the reactant that will be completely consumed first.
To calculate the moles of MgSO4 produced, we can use the mole ratio between Mg and MgSO4:
Moles of MgSO4 = 0.099 mol Mg * (1 mol MgSO4 / 1 mol Mg) = 0.099 mol
To calculate the mass of water produced when the H2 gas reacts with O2, we need to use the balanced equation for the reaction:
H2 + O2 -> H2O
From the equation, we can see that 1 mole of H2 reacts with 1 mole of O2 to produce 2 moles of H2O. Since we know that the moles of H2 gas produced is 0.099 mol, we can use the mole ratio to calculate the moles of water produced:
Moles of H2O = 0.099 mol H2 * (2 mol H2O / 1 mol H2) = 0.198 mol
To calculate the mass of water produced, we can use the molar mass of water (18.015 g/mol):
Mass of water = 0.198 mol * 18.015 g/mol = 3.561 g
The mass of excess reactant can be calculated by subtracting the mass of limiting reactant consumed from the original mass of excess reactant. Since the limiting reactant is Mg, we can calculate the moles of H2SO4 used:
Moles of H2SO4 = 0.099 mol Mg * (1 mol H2SO4 / 1 mol Mg) = 0.099 mol
To calculate the mass of excess H2SO4, we can use the molar mass of H2SO4 (98.09 g/mol):
Mass of excess H2SO4 = 0.099 mol * 98.09 g/mol = 9.72 g