Answer to Question #339728 in Physical Chemistry for Luke

Question

In the reaction Mg(OH)2 + 2HCl → MgCl2 + 2H2 O, you were given 15 grams of Mg(OH)2 and 25 grams of HCl. Which of the two reactants is the limiting reagent? How much MgCl2 will be produced in the reaction? (Mg(OH)2 = 58.32 g/mol, HCl = 36.46 g/mol, MgCl2 = 95.32 g/mol)

Answer 1

Answer:

Step 1: Calculate the molar mass of each reactant.

For Mg(OH)₂

molar mass = (24.31 g/mol × 1) + (16.00 g/mol × 2) + (1.008 g/mol × 2)

molar mass = 58.326 g/mol

For HCl

molar mass = (1.008 g/mol × 1) + (35.45 g/mol × 1)

molar mass = 36.458 g/mol

Step 2: Determine the mole ratios needed.

Using Mg(OH)₂

mole ratio = 1 mol Mg(OH)₂ : 1 mol MgCl₂

Using HCl

mole ratio = 2 mol HCl : 1 mol MgCl₂

Step 3: Calculate the number of moles of MgCl₂ produced using the given masses of two reactants and their mole ratios.

Using Mg(OH)₂

\text{moles of MgCl₂ = 16.0 g Mg(OH)₂} × \frac{\text{1 mol Mg(OH)₂}}{\text{58.326 g Mg(OH)₂}} × \frac{\text{1 mol MgCl₂}}{\text{1 mol Mg(OH)₂}}moles of MgCl₂ = 16.0 g Mg(OH)₂×58.326 g Mg(OH)₂1 mol Mg(OH)₂×1 mol Mg(OH)₂1 mol MgCl₂

\text{moles of MgCl₂ = 0.274 mol}moles of MgCl₂ = 0.274 mol

Using HCl

\text{moles of MgCl₂ = 11.0 g HCl} × \frac{\text{1 mol HCl}}{\text{36.458 g HCl}} × \frac{\text{1 mol MgCl₂}}{\text{2 mol HCl}}moles of MgCl₂ = 11.0 g HCl×36.458 g HCl1 mol HCl×2 mol HCl1 mol MgCl₂

\text{moles of MgCl₂ = 0.15086 mol}moles of MgCl₂ = 0.15086 mol

Step 4: Determine the limiting reactant.

Since HCl produces less number of moles of MgCl₂

\boxed{\text{HCl is the limiting reactant.}}HCl is the limiting reactant.

Solution: (b)

Step 1: Calculate the molar mass of MgCl₂.

molar mass = (24.31 g/mol × 1) + (35.45 g/mol × 2)

molar mass = 95.21 g/mol

Step 2: Calculate the mass of MgCl₂ produced.

Since HCl is the limiting reactant

\text{mass of MgCl₂ = 0.15086 mol MgCl₂} × \frac{\text{95.21 g MgCl₂}}{\text{1 mol MgCl₂}}mass of MgCl₂ = 0.15086 mol MgCl₂×1 mol MgCl₂95.21 g MgCl₂

\boxed{\text{mass of MgCl₂ = 14.4 g}}mass of MgCl₂ = 14.4 g