Answer:
To determine the mass of water produced when 45g of HCl reacts with 27g of MgO, we need to balance the chemical equation for the reaction and calculate the stoichiometry.
The balanced chemical equation for the reaction between HCl and MgO is:
2 HCl + MgO -> MgCl₂ + H₂O
From the balanced equation, we can see that the stoichiometric ratio between HCl and water is 2:1. This means that for every 2 moles of HCl used, 1 mole of water is produced.
First, we need to determine the number of moles of HCl and MgO available:
Moles of HCl = Mass of HCl / Molar mass of HCl
Moles of HCl = 45g / 36.46 g/mol ≈ 1.233 mol
Moles of MgO = Mass of MgO / Molar mass of MgO
Moles of MgO = 27g / 40.31 g/mol ≈ 0.670 mol
Next, we determine the limiting reactant by comparing the moles of HCl and MgO. The reactant that produces the smallest amount of product is the limiting reactant.
Since the stoichiometric ratio between HCl and water is 2:1, we can calculate the maximum moles of water that can be formed using the moles of HCl:
Moles of water = 1.233 mol of HCl / 2 = 0.617 mol
Now, we can calculate the mass of water produced using the molar mass of water:
Mass of water = Moles of water × Molar mass of water
Mass of water = 0.617 mol × 18.015 g/mol ≈ 11.119 g
Therefore, the expected mass of water produced is approximately 11.119 grams.
To calculate the percent yield, we compare the actual mass of water produced (9.5g) to the expected mass (11.119g) and use the formula:
Percent yield = (actual mass / expected mass) × 100%
Percent yield = (9.5g / 11.119g) × 100% ≈ 85.48%
Therefore, the percent yield of the reaction is approximately 85.48%.
