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# Answer to Question #195119 in Chemistry for Dan

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.

1
2021-05-19T02:28:30-0400

Solution:

Balanced chemical equation:

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

a)

Mass of H2SO4 = (Mass of H2SO4 solution × Mass percent) / 100%

Mass of H2SO4 = (500 g × 46%) / 100 % = 230 g

The molar mass of H2SO4 is 98.079 g/mol.

Hence,

(230 g H2SO4) × (1 mol H2SO4 / 98.079 g H2SO4) = 2.345 mol H2SO4

b)

(Moles of Al) / 2 = (Moles of H2SO4) / 3 (according to the equation above)

Moles of Al = 2 × Moles of H2SO4 / 3 = (2 × 2.345 mol) / 3 = 1.563 mol Al

c)

Moles of Al2(SO4)3 = (Moles of H2SO4) / 3 (according to the equation above)

Moles of Al2(SO4)3 = 2.345 mol / 3 = 0.782 mol Al2(SO4)3

The molar mass of Al2(SO4)3 is 342.15 g/mol.

Hence,

(0.782 mol Al2(SO4)3) × (342.15 g Al2(SO4)3 / 1 mol Al2(SO4)3) = 267.56 g Al2(SO4)3

Balanced chemical equation: 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3H2(g)

a) 2.345 mol H2SO4

b) 1.563 mol Al

c) 0.782 mol Al2(SO4)3 or 267.56 g Al2(SO4)3

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