Question

1. in the reaction of 2k₃po₄ + 3h₂so₄ → 2h₃po₄ + 3k₂so₄, 15.32 g of h₂so₄ are reacted with an unlimited amount of potassium phosphate.

a. what is the theoretical yield, in grams, of k₂so₄?
b. what is the percent yield of k₂so₄ if only 18.99 g are produced in lab?

Explanation:

Step1: Calculate molar masses

The molar mass of $H_2SO_4$ is $M_{H_2SO_4}=2\times1 + 32+4\times16=98\ g/mol$. The molar mass of $K_2SO_4$ is $M_{K_2SO_4}=2\times39 + 32+4\times16 = 174\ g/mol$.

Step2: Determine moles of $H_2SO_4$

The number of moles of $H_2SO_4$, $n_{H_2SO_4}=\frac{m_{H_2SO_4}}{M_{H_2SO_4}}=\frac{15.32\ g}{98\ g/mol}=0.1563\ mol$.

Step3: Use mole - ratio to find moles of $K_2SO_4$

From the balanced chemical equation $2K_3PO_4 + 3H_2SO_4
ightarrow2H_3PO_4 + 3K_2SO_4$, the mole - ratio of $H_2SO_4$ to $K_2SO_4$ is 3:3 = 1:1. So, $n_{K_2SO_4}=n_{H_2SO_4}=0.1563\ mol$.

Step4: Calculate theoretical yield of $K_2SO_4$

The theoretical mass of $K_2SO_4$, $m_{theo}=n_{K_2SO_4}\times M_{K_2SO_4}=0.1563\ mol\times174\ g/mol = 27.1962\ g$.

Step5: Calculate percent yield

The percent yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$. Given actual yield $m_{actual}=18.99\ g$ and theoretical yield $m_{theo}=27.1962\ g$. So, $\text{Percent Yield}=\frac{18.99\ g}{27.1962\ g}\times100\%\approx69.83\%$.

Answer:

a. $27.20\ g$
b. $69.83\%$