# 0.495g of Al foil and dissolves it w/ 25.3mL of 1.4M of KOH. After cooling, the student adds 11.9 mL of 9.0 M H2SO4. What is the molar mass of alum?

Feb 17, 2015

In order for you to actually calculate the molar mass of alum, a mass produced by the reaction should have been given to you. Since molar mass is defined as the mass of 1 mole of a substance, you'd need both the number of moles of alum - which you can easily calculate - and the mass produced - which you do not have.

There are actually four reactions that take place here (I won't go into the net ionic equations)

$2 A l \left(s\right) + 2 K O H \left(a q\right) + 6 {H}_{2} O \left(l\right) \to 2 K A l {\left(O H\right)}_{4} \left(a q\right) + 3 {H}_{2} \left(g\right)$

$2 K A l {\left(O H\right)}_{4} \left(a q\right) + {H}_{2} S {O}_{4} \left(a q\right) \to 2 A l {\left(O H\right)}_{3} \left(s\right) + {K}_{2} S {O}_{4} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

$2 A l {\left(O H\right)}_{3} \left(s\right) + {H}_{2} S {O}_{4} \to A {l}_{2} {\left(S {O}_{4}\right)}_{3} \left(a q\right) + 6 {H}_{2} O \left(l\right)$

${K}_{2} S {O}_{4} \left(a q\right) + A {l}_{2} {\left(S {O}_{4}\right)}_{3} \left(a q\right) + 24 {H}_{2} O \left(l\right) \to 2 K A l {\left(S {O}_{4}\right)}_{2} \cdot 12 {H}_{2} O \left(s\right)$

If you add all these reactions together, eliminate the compounds that are both on the reactants' and on the products' side, and balance all the atoms you'll get the overall complete reaction

$2 A l \left(s\right) + 2 K O H \left(a q\right) + 4 {H}_{2} S {O}_{4} \left(a q\right) + 22 {H}_{2} O \left(l\right) \to 2 K A l {\left(S {O}_{4}\right)}_{2} \cdot 12 {H}_{2} O \left(s\right) + 3 {H}_{2} \left(g\right)$

Now, the important thing to notice here is that you have a $\text{1:1}$ mole ratio between aluminium and alum - 2 moles of aluminium will produce 2 moles of alum.

Before figuring out how many moles of alum are produced by the reaction, you need to determine whether or not you have a limiting reagent, i.e if you have insufficient moles of $K O H$ or ${H}_{2} S {O}_{4}$.

Determine the number of moles of aluminium, potassium hydroxide, and sulfuric acid

$\text{0.495 g Al" * ("1 mole")/("27.0 g") = "0.0183 moles Al}$

C = n/V => n_("KOH") = C * V = 25.3 * 10^(-3)"L" * "1.4 M" = "0.0354 moles KOH"

${n}_{{H}_{2} S {O}_{4}} = C \cdot V = 11.9 \cdot {10}^{- 3} \text{L" * "9.0 M" = "0.107 moles}$ ${H}_{2} S {O}_{4}$

You need to examine the mole ratio aluminium has with both $K O H$ and ${H}_{2} S {O}_{4}$ in order to determine how many moles actually react.

So, 2 moles of aluminium need 2 moles of $K O H$ and 4 moles of ${H}_{2} S {O}_{4}$. If you look at how many moles of each you have, you'll notice that aluminium acts as a limiting reagent, which means that all the aluminium will react and you'll have excess $K O H$ and ${H}_{2} S {O}_{4}$.

Therefore, the number of moles of alum produced will be

$\text{0.0183 moles Al" * ("1 mole alum")/("1 mole Al") = "0.0183 moles alum}$

You could use the molar masses of all species, except for alum, to determine the total mass of the reactants and the total mass of the products. This will get you the mass of alum produced, which you can then use to determine its molar mass.

The underlying principle here is that you have conservation of mass. If you know what mass reacted, you know what mass is produced.

Now, I could calculate all those masses and eventually determine how much alum is produced, but that would make a long answer even longer, and I'm not sure that's what you have to do.

The simple way to calculate the molar mass of alum is to add the molar masses of each atom that forms the compound. For alum, you'd get

${M}_{M} = 39.1 + 27 + 2 \cdot 32 + 8 \cdot 16 + 12 \cdot \left(2 + 16\right)$

${M}_{M} = \text{474.1 g/mol}$

So maybe all you had to do was determine the formula for alum, which would make almost everything I wrote here unnecessary.