Question

Determine the theoretical yield of alum (MM=474.38), to two decimal places, if 0.49 g of aluminum (MM = 26.98 g/mol) are reacted.

Answer 1

The theoretical yield of alum = 8.62 g

Explanation

Given:

Mass of aluminum that reacted = 0.49 g

Molar mass of aluminum = 26.98 g/mol

Molecular mass of alum = 474.38 g/mol

What to find:

The theoretical yield of alum.

Explanation:

If 0.46 g of aluminum reacted, then the number of moles of aluminum that reacted will be:

`\begin{gathered} \text{moles of aluminum reacted }=\frac{reacing\text{ }mass\text{ of aluminum}}{molar\text{ mass of aluminum}}=\frac{0.49\text{ }g}{26.98\text{ g/mol}} \\ \text{moles of aluminum reacted }=0.018161601\text{ mol} \end{gathered}`

The molecular mass of alum = 474.38 g/mol

The chemical formula of alum is KAl(SO₄)₂.12H₂O

1 mol Al → KAl(SO₄)₂.12H₂O

Thus, the theoretical yield of alum is:

`\begin{gathered} \text{Theoretical yield of alum }=moles\text{ Al that reacted }* Molecular\text{ weight of alum} \\ \text{Theoretical yield of alum }=0.018161601\text{mol }*474.38\text{ g/mol} \\ \text{Theoretical yield of alum }=8.615500\text{ g} \\ To\text{ two decimal places,} \\ \text{Theoretical yield of alum }\approx8.62\text{ g} \end{gathered}`