Question

1. 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g) a. Determine the volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum to produce aluminum sulfate. b. Determine the % yield if 112 g of aluminum sulfate is produced under the above conditions.

Answer 1

Answer:

a. 167 mL b. 39.27 %

Explanation:

a. From the chemical equation. 2 mole of Al reacts with 3 mole H₂SO₄ to produce 1 mol  Al₂(SO₄)₃.

Now, we calculate the number of moles of Al in 45.0 g Al.

We know number of moles, n = m/M where m = mass of Al = 45.0 g and M = molar mass of Al = 26.98 g/mol.

So n = 45.0 g/26.98 g/mol = 1.668 mol

Since 2 mole of Al reacts with 3 mole H₂SO₄, then 1.668 mole of Al reacts with x mole H₂SO₄. So, x = 3 × 1.668/2 mol = 2.5 mol

So, we have 2.5 mol H₂SO₄.

Now number of moles of H₂SO₄, n = CV where C = concentration of H₂SO₄ = 15.0 M = 15.0 mol/L and V = volume of H₂SO₄.

V = n/C

= 2.5 mol/15.0 mol/L

= 0.167 L

= 167 mL of 15.0 M H₂SO₄ reacts with 45.0 g Al to produce aluminum sulfate.

b. From the chemical reaction, 2 mol Al produces 1 mol Al₂(SO₄)₃

Therefore 1.668 mol Al will produce x mol  Al₂(SO₄)₃. So, x = 1 mol × 1.668 mol/2 mol = 0.834 mol

So, we need to find the mass of 0.834 mol  Al₂(SO₄)₃. Now molar mass  Al₂(SO₄)₃ = 2 × 26.98 g/mol + 3 × 32 g/mol + 4 × 3 × 16 g/mol = 53.96 g/mol + 96 g/mol + 192 g/mol = 341.96 g/mol.

Also number of moles of  Al₂(SO₄)₃, n = mass of  Al₂(SO₄)₃,m/molar mass  Al₂(SO₄)₃, M

n =m/M

So, m = nM = 0.834 mol × 341.96 g/mol = 285.2 g

% yield = Actual yield/theoretical yield × 100 %

Actual yield = 112 g, /theoretical yield = 285.2 g

So, % yield = 112 g/285.2 g × 100 %

= 0.3927 × 100 %

=  39.27 %

Answer 2

The volume (mL) of 15.0 M sulfuric acid needed to react with 45.0 g of aluminum is 166mL and % yield of the reaction is 39.46%.

How do we calculate moles?

Moles of any substance will be calculated by using the below formula as:
n = W/M, where

W = given mass

M = molar mass

Given chemical reaction is :

2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

Moles of 45g of Al will be calculated as:

n = 45g / 27g/mol = 1.66 mole

From the stoichiometry of the reaction, it is clear that:

1.66 moles of Al = react with 3/2×1.66=2.49 moles of H₂SO₄

By using the formula of molarity we can calculate the volume of H₂SO₄ as:

M = n/V

V = (2.49) / (15) = 0.166L = 166mL

Again from the stoichiometry it is clear that:
1.66 moles of Al = produces 1/2×1.66= 0.83 moles of Al₂(SO₄)₃

Mass of 0.83 moles of Al₂(SO₄)₃ = (0.83mol)(341.96g/mol) = 283.82 g

Given actual yield of Al₂(SO₄)₃ = 112g

% yield will be calculated as:

Percent yield = (Actual yield/Theoretical yield) × 100

% yield = (112/283.82) × 100 = 39.46%

Hence required values are discussed above.

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