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.

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

We need 79.6grams of H2SO4

Explanation:

Step 1: data given

Mass of aluminium = 14.6 grams

Molar mass aluminium = 26.98 g/mol

Molar mass H2SO4 = 98.08 g/mol

Step 2: The balanced equation

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

Step 3: Calculate moles Aluminium

Moles Aluminium = mass aluminium / molar mass aluminium

Moles aluminium = 14.6 grams / 26.98 g/mol

Moles aluminium = 0.541 moles

Step 4: Calculate moles H2SO4 need

For 2 moles Al we need 3 moles H2SO4

For 0.541 moles Al we need 3/2 * 0.541 = 0.8115 moles H2SO4

Step 5: Calculate mass H2SO4

Mass H2SO4 = 0.8115 moles * 98.08 g/mol

Mass H2SO4 = 79.6 grams

We need 79.6grams of H2SO4