Question

1. _8_ mg + _2_ hcl → _2_ mgcl₂ + _1_ h₂\

3a what mass of hcl is consumed by the reaction of 2.5 moles magnesium?\
3b what mass of mgcl₂ is produced if 3.67 moles of hcl react?\
3c how many moles of hydrogen gas is produced when 1.7 moles of magnesium reacts with excess hcl?

Explanation:

3a

Step1: Balance the chemical equation

The given equation is \( 8\text{Mg} + 2\text{HCl}
ightarrow 2\text{MgCl}_2 + 1\text{H}_2 \). Simplify the coefficients by dividing by 2: \( 4\text{Mg} + \text{HCl}
ightarrow \text{MgCl}_2 + \frac{1}{2}\text{H}_2 \). Wait, actually, the correct balancing should be: \( \text{Mg} + 2\text{HCl}
ightarrow \text{MgCl}_2 + \text{H}_2 \). Let's check the original given coefficients: 8 Mg, 2 HCl, 2 MgCl₂, 1 H₂. To balance Mg: 8 on left, 2 on right? No, that's incorrect. The correct balanced equation for Mg and HCl reaction is \( \text{Mg} + 2\text{HCl}
ightarrow \text{MgCl}_2 + \text{H}_2 \). So the mole ratio of Mg to HCl is 1:2.

Step2: Determine moles of HCl

Given moles of Mg = 2.5 mol. From the balanced equation (correcting the given one, assuming a typo, the correct ratio is 1 mol Mg reacts with 2 mol HCl). So moles of HCl = 2.5 mol Mg × \( \frac{2\text{ mol HCl}}{1\text{ mol Mg}} \) = 5 mol HCl.

Step3: Calculate mass of HCl

Molar mass of HCl: H = 1 g/mol, Cl = 35.45 g/mol, so molar mass = 1 + 35.45 = 36.45 g/mol. Mass = moles × molar mass = 5 mol × 36.45 g/mol = 182.25 g.

Wait, but the given equation has 8 Mg, 2 HCl, 2 MgCl₂, 1 H₂. Let's use the given coefficients. Mole ratio of Mg to HCl: 8 mol Mg : 2 mol HCl = 4:1. So moles of HCl = 2.5 mol Mg × \( \frac{2\text{ mol HCl}}{8\text{ mol Mg}} \) = 2.5 × \( \frac{1}{4} \) = 0.625 mol HCl? No, that can't be. The original equation is incorrectly balanced. The correct reaction is Mg + 2 HCl → MgCl₂ + H₂. So we should use the correct balanced equation. Let's proceed with the correct balancing.

Correct balanced equation: \( \text{Mg} + 2\text{HCl}
ightarrow \text{MgCl}_2 + \text{H}_2 \) (mole ratio Mg:HCl = 1:2).

So moles of HCl = 2.5 mol Mg × 2 = 5 mol HCl.

Mass of HCl = 5 mol × 36.45 g/mol = 182.25 g.

Step1: Balance the equation (correctly: \( \text{Mg} + 2\text{HCl}

ightarrow \text{MgCl}_2 + \text{H}_2 \))
Mole ratio of HCl to MgCl₂: 2 mol HCl : 1 mol MgCl₂.

Step2: Determine moles of MgCl₂

Given moles of HCl = 3.67 mol. Moles of MgCl₂ = 3.67 mol HCl × \( \frac{1\text{ mol MgCl}_2}{2\text{ mol HCl}} \) = 1.835 mol MgCl₂.

Step3: Calculate molar mass of MgCl₂

Mg = 24.305 g/mol, Cl = 35.45 g/mol. Molar mass = 24.305 + 2×35.45 = 24.305 + 70.9 = 95.205 g/mol.

Step4: Calculate mass of MgCl₂

Mass = 1.835 mol × 95.205 g/mol ≈ 174.7 g.

Using the given (incorrect) equation: 8 Mg + 2 HCl → 2 MgCl₂ + H₂. Mole ratio HCl:MgCl₂ = 2:2 = 1:1. So moles of MgCl₂ = 3.67 mol HCl × \( \frac{2\text{ mol MgCl}_2}{2\text{ mol HCl}} \) = 3.67 mol. Molar mass of MgCl₂ = 95.205 g/mol. Mass = 3.67 mol × 95.205 g/mol ≈ 349.4 g. But the correct balanced equation is different. Assuming the given equation is correct as written (8 Mg, 2 HCl, 2 MgCl₂, 1 H₂), then:

Mole ratio HCl:MgCl₂ = 2:2 = 1:1. So moles of MgCl₂ = 3.67 mol. Mass = 3.67 mol × 95.205 g/mol ≈ 349 g.

Step1: Balance the equation (correct: \( \text{Mg} + 2\text{HCl}

ightarrow \text{MgCl}_2 + \text{H}_2 \))
Mole ratio Mg:H₂ = 1:1.

Using given equation: 8 Mg + 2 HCl → 2 MgCl₂ + 1 H₂. Mole ratio Mg:H₂ = 8:1.

Step2: Calculate moles of H₂

Given moles of Mg = 1.7 mol. Using given equation: moles of H₂ = 1.7 mol Mg × \( \frac{1\text{ mol H}_2}{8\text{ mol Mg}} \) = 0.2125 mol.

Using correct equation: moles of H₂ = 1.7 mol Mg × \( \frac{1\text{ mol H}_2}{1\text{ mol Mg}} \) = 1.7 mol.

Assuming the given equation is correct as written (8 Mg, 2 HCl, 2 MgCl₂, 1 H₂), then:

Moles of H₂ = 1.7 mol Mg × (1 mol H₂ / 8 mol Mg) = 0.2125 mol ≈ 0.213 mol.

Answer:

182.25 g (or considering significant figures, 180 g or 182 g depending on precision)

3b