## 9.0 gm Al, 12.0 gm sulphur and 5.6 L ${O}_{2}$ gas (at ${0}^{\circ} C$ and 1 atm) are allowed to react. The maximum mass of $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$, which may form is - 29.0 gm 57.0 gm 42.75 gm 14.25 gm

Apr 10, 2018

#### Explanation:

1) Write a balanced chemical reaction equation.

$2 A l + 3 S + 6 {O}_{2} \rightarrow A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

2) Determine the number of moles of each reactant.

${n}_{A l}^{0} =$9 gm Al(1 mol Al/26.98 gm Al) = 0.334 mol Al

${n}_{S}^{0} =$12 gm S(1mol S/32.06 gm S) = 0.374 mol S

For the diatomic oxygen, we must use the ideal gas law.

${n}_{{O}_{2}}^{o} = \frac{P V}{R T} = \frac{1 \cdot 5.6}{0.082057 \cdot 273} = 0.25$ mol ${O}_{2}$

3) Determine the limiting reagent.

The number of starting moles of each reactant are comparable, however we require twice as many moles of diatomic oxygen than sulfur and three times as many moles of diatomic oxygen than aluminum, so this suggests that ${O}_{2}$ is the limiting reagent.

4) Calculate the molecular weight of the product

Molecular Weight of $A {l}_{2} {\left(S O 4\right)}_{3}$

=2(26.98)+3(32.06)+12(16)=$342.1$ gm $A {l}_{2} {\left(S O 4\right)}_{3}$ per mole

5) Calculate the number of grams of product that can be produced from the limiting reagent.

0.25 mol ${O}_{2} \times$ (1 mol $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$)/(6 mol ${O}_{2}$)$\times$(342.1 gm $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$)/(mole $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$)=14.25 gm $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$