Question #195119

Sulphuric acid (H2SO4) reacts with aluminium to produce aluminium sulphate

(Al2 (SO4)3 ) and hydrogen (gas). Write and balance the reaction. If 500 g of a 46% by

mass solution of sulphuric acid is combined with aluminium

a) How many moles of sulphuric acid will there be in 500 g of 46% solution of this

acid?

b) Calculate the moles of aluminium necessary to completely consume the moles of

sulphuric acid obtained in the previous section.

c) Determine the moles of aluminium sulphate that will be produced and express

this quantity in grams.

Expert's answer

**Solution:**

Balanced chemical equation:

2Al(s) + 3H_{2}SO_{4}(aq) → Al_{2}(SO_{4})_{3}(aq) + 3H_{2}(g)

**a)**

Mass of H_{2}SO_{4} = (Mass of H_{2}SO_{4} solution × Mass percent) / 100%

Mass of H_{2}SO_{4} = (500 g × 46%) / 100 % = 230 g

The molar mass of H_{2}SO_{4} is 98.079 g/mol.

Hence,

(230 g H_{2}SO_{4}) × (1 mol H_{2}SO_{4} / 98.079 g H_{2}SO_{4}) = **2.345 mol H**_{2}**SO**_{4}

**b)**

(Moles of Al) / 2 = (Moles of H_{2}SO_{4})_{ }/ 3 (according to the equation above)

Moles of Al = 2 × Moles of H_{2}SO_{4 }/ 3 = (2 × 2.345 mol) / 3 = **1.563 mol Al**

**c)**

Moles of Al_{2}(SO_{4})_{3} = (Moles of H_{2}SO_{4})_{ }/ 3 (according to the equation above)

Moles of Al_{2}(SO_{4})_{3} = 2.345 mol / 3 = **0.782 mol Al**_{2}**(SO**_{4}**)**_{3}

The molar mass of Al_{2}(SO_{4})_{3} is 342.15 g/mol.

Hence,

(0.782 mol Al_{2}(SO_{4})_{3}) × (342.15 g Al_{2}(SO_{4})_{3} / 1 mol Al_{2}(SO_{4})_{3}) = **267.56 g Al**_{2}**(SO**_{4}**)**_{3}

**Answers:**

**Balanced chemical equation: 2Al(s) + 3H**_{2}**SO**_{4}**(aq) → Al**_{2}**(SO**_{4}**)**_{3}**(aq) + 3H**_{2}**(g)**

**a) 2.345 mol H**_{2}**SO**_{4}

**b) 1.563 mol Al**

**c) 0.782 mol Al**_{2}**(SO**_{4}**)**_{3}** or 267.56 g Al**_{2}**(SO**_{4}**)**_{3}

Learn more about our help with Assignments: Chemistry

## Comments

## Leave a comment