Question

What is the percent yield of aluminum phosphate if a solution containing 33.4 g of sodium phosphate produced 19.6 g of aluminum phosphate when reacted with excess aluminum chloride in solution

Answer 1

Answer: The percent yield of aluminium phosphate in the reaction is 78.78 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For sodium phosphate:

Given mass of sodium phosphate = 33.4 g

Molar mass of sodium phosphate = 164 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium phosphate}=\frac{33.4g}{164g/mol}=0.204mol$

The chemical equation for the reaction of sodium phosphate and aluminium chloride is:

$Na_3PO_4+AlCl_3\rightarrow 3NaCl+AlPO_4$

By Stoichiometry of the reaction:

1 mole of sodium phosphate produces 1 mole of aluminium phosphate

So, 0.204 moles of sodium phosphate will produce = $\frac{1}{1}\times 0.204=0.204$ moles of aluminium phosphate

• Now, calculating the mass of aluminium phosphate from equation 1, we get:

Molar mass of aluminium phosphate = 122 g/mol

Moles of aluminium phosphate = 0.204 moles

Putting values in equation 1, we get:

$0.204mol=\frac{\text{Mass of aluminium phosphate}}{122g/mol}\\\\\text{Mass of aluminium phosphate}=(0.204mol\times 122g/mol)=24.88g$

• To calculate the percentage yield of aluminium phosphate, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of aluminium phosphate = 19.6 g

Theoretical yield of aluminium phosphate = 24.88 g

Putting values in above equation, we get:

$\%\text{ yield of aluminium phosphate}=\frac{19.6g}{24.88g}\times 100\\\\\% \text{yield of aluminium phosphate}=78.78\%$

Hence, the percent yield of aluminium phosphate in the reaction is 78.78 %