Question

Sulfuric acid dissolves aluminum metal according to the following reaction:

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

Suppose you wanted to dissolve an aluminum block with a mass of 15.3 g .

Part A:

What minimum mass of H2SO4 would you need?

Express your answer in grams.

Part B:

What mass of H2 gas would be produced by the complete reaction of the aluminum block?

Express your answer in grams.

Answer 1

A)

Molar mass of Al = 26.98 g/mol

mass of Al = 15.3 g

mol of Al = (mass)/(molar mass)

= 15.3/26.98

= 0.5671 mol

According to balanced equation

mol of H2SO4 required = (3/2)* moles of Al

= (3/2)*0.5671

= 0.8506 mol

Molar mass of H2SO4,

MM = 2*MM(H) + 1*MM(S) + 4*MM(O)

= 2*1.008 + 1*32.07 + 4*16.0

= 98.086 g/mol

mass of H2SO4 = number of mol * molar mass

= 0.8506*98.09

= 83.43 g

Answer: 83.4 g

B)

According to balanced equation

mol of H2 formed = (3/2)* moles of Al

= (3/2)*0.5671

= 0.8506 mol

Molar mass of H2 = 2.016 g/mol

mass of H2 = number of mol * molar mass

= 0.8506*2.016

= 1.715 g

Answer: 1.72 g