Question

2. Given the following reaction involving 10.6 g of Al: Al(SO4)3 (s) + 3 H2(g) 2 Al(s) + 3 H2SO, (aq) - a) How many mol of Al are used? Ans. b) How many mol of Al2(SO4)3 (s) could be produced by the reaction with excess H SO? Ans.

Answer 1

a) Number of moles of Al = 10.6 g / 26.9815 g/mol = 0.393 mole

Therefore, the number of moles of Al used = 0.393 mole

b)

From the balanced equation we can say that

2 mole of Al produces 1 mole of Al2(SO4)3 so

0.393 mole of Al will produce

= 0.393 mole of Al *(1 mole of Al2(SO4)3 / 2 mole of Al)

= 0.197 mole of Al2(SO4)3

Therefore, the number of moles of Al2(SO4)3 produced would be 0.197