Question

Sulfuric acid dissolves aluminum metal according to the following reaction: 2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g)2Al(s)+3H2SO4(aq)→Al2(SO4)3(aq)+3H2(g) Suppose you wanted to dissolve an aluminum block with a mass of 14.6 gg . Part A What minimum mass of H2SO4H2SO4 would you need? Express your answer in grams.

Answer 1

Answer:

76.3 g of H₂SO₄ are needed in this reaction

Explanation:

Let's begin with the reaction:

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

Ratio in the reactants side is 2:3. It means that 2 moles of aluminum need 3 moles of sulfuric acid to react.

We determine the mass of Al → 14.6 g . 1mol / 26.98 g = 0.519 moles

Let's propose this rule of three:

2 moles of Al react with 3 moles of H₂SO₄

Then 0.519 moles of Al will react with (0.519 . 3) /2 = 0.778 moles of acid.

We convert the moles to mass, to find the answer:

0.778 mol . 98 g / 1mol = 76.3 g

Answer 2

The minimum mass of H₂SO₄ that would be needed is 79.6 g

From the question,

We are to determine the minimum mass of H₂SO₄ required to dissolve 14.6 g of aluminum block.

The given balanced chemical equation for the reaction is

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

This means,

3 moles of sulfuric acid is required to dissolve 2 moles of Aluminum

First, we will determine the number of moles of aluminum present

From the given information,

Mass of aluminum block = 14.6 g

Using the formula

$Number\ of\ moles = \frac{Mass}{Atomic\ mass}$

Atomic mass of aluminum = 26.98 g/mol

∴ Number of moles of Al present = $\frac{14.6}{26.98}$

Number of moles of Al present = 0.54114 mole

Since

3 moles of sulfuric acid is required to dissolve 2 moles of Aluminum

Then,

x moles of sulfuric acid is required to dissolve 0.54114 mole of Aluminum

x = $\frac{3\times 0.54114}{2}$

x = 0.81171 mole

∴ The minimum number of mole of sulfuric acid required is 0.81171 mole

Now, for the minimum mass of sulfuric acid required

Using the formula

Mass = Number of moles × Molar mass

Molar mass of sulfuric acid = 98.079 g/mol

∴ Minimum Mass of sulfuric acid required = 0.81171 × 98.079

Minimum Mass of sulfuric acid required = 79.6117 g

Minimum Mass of sulfuric acid required ≅ 79.6 g

Hence, the minimum mass of H₂SO₄ that would be needed is 79.6 g

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