Question

a reaction was conducted between barium nitrate and sodium phosphate.
3 ba(no₃)₂ + 2 na₃po₄ → 6 nano₃ + ba₃(po₄)₂
what is the percent yield if 0.3 mol ba(no₃)₂ and 0.25 mol na₃po₄ react to produce 0.095 mol ba₃(po₄)₂?
view available hint(s)
76%
19%
38%
95%

Explanation:

Step1: Determine the limiting reactant

The mole - ratio of $Ba(NO_3)_2$ to $Na_3PO_4$ from the balanced equation is $\frac{3}{2}$.
For $0.3$ mol of $Ba(NO_3)_2$, the moles of $Na_3PO_4$ required is $n_{Na_3PO_4}=\frac{2}{3}\times0.3 = 0.2$ mol. Since we have $0.25$ mol of $Na_3PO_4$, $Ba(NO_3)_2$ is the limiting reactant.

Step2: Calculate the theoretical yield of $Ba_3(PO_4)_2$

From the balanced equation, the mole - ratio of $Ba(NO_3)_2$ to $Ba_3(PO_4)_2$ is $\frac{3}{1}$.
If $n_{Ba(NO_3)_2}=0.3$ mol, then the theoretical yield of $Ba_3(PO_4)_2$, $n_{Ba_3(PO_4)_2}^{theo}=\frac{1}{3}\times0.3 = 0.1$ mol.

Step3: Calculate the percent yield

The percent - yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$.
Given actual yield $n_{Ba_3(PO_4)_2}^{act}=0.095$ mol and theoretical yield $n_{Ba_3(PO_4)_2}^{theo}=0.1$ mol.
$\text{Percent Yield}=\frac{0.095}{0.1}\times100\% = 95\%$

Answer:

$95\%$