# Aluminum will react with sulfuric acid in the following reaction: 2Al(s) + 3H_2SO_4(l) -> Al_2(SO_4)_3(aq) + 3H_2(g). How many moles of H_2SO_4 will react with 18 mol Al? How many moles of each product will be produced?

Jan 4, 2018

$= 27 m o l {H}_{2} S {O}_{4}$

#### Explanation:

With reference to the given balanced equation for the mole ratios, the needed number of moles for each remaining compound involved in the reaction using the mass of $A l$ as basis for the molar conversions can be computed as;

$\eta {H}_{2} S {O}_{4} :$
$= 18 \cancel{m o l A l} \times \frac{3 m o l {H}_{2} S {O}_{4}}{2 \cancel{m o l A l}}$
$= 27 m o l {H}_{2} S {O}_{4}$

$\eta A {l}_{2} {\left(S {O}_{4}\right)}_{3} :$
$= 18 \cancel{m o l A l} \times \frac{1 m o l A {l}_{2} {\left(S {O}_{4}\right)}_{3}}{2 \cancel{m o l A l}}$
$= 9 m o l A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

$\eta {H}_{2} :$
$= 18 \cancel{m o l A l} \times \frac{3 m o l {H}_{2}}{2 \cancel{m o l A l}}$
$= 27 m o l {H}_{2}$