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 %

Step-by-step 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}=(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 =
`(1)/(1)* 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* 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}}* 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}=(19.6g)/(24.88g)* 100\\\\\% \text{yield of aluminium phosphate}=78.78\%`

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