Question

of pbcl₂ produced from complete consumption of pb(no₃)₂: theoretical yield = mol pbcl₂ × molar mass of pbcl₂ = 1.26×10⁻² mol pbcl₂ × 278.106 g pbcl₂ / 1 mol pbcl₂ = 3.51 g pbcl₂ part c determine the percent yield. express your answer to three significant figures.

Explanation:

Step1: Recall percent - yield formula

Percent yield = $\frac{\text{actual yield}}{\text{theoretical yield}}\times100\%$
Since the actual yield is not given in the problem statement, assume we are given an actual yield value of $a$ grams (for the sake of showing the process). We know the theoretical yield of $PbCl_2$ is $3.51$ g from the previous part.

Step2: Calculate percent yield

Percent yield = $\frac{a}{3.51}\times100\%$

Let's assume the actual yield is $2.80$ g.
Percent yield = $\frac{2.80}{3.51}\times100\%$
Percent yield = $0.798\times100\% = 79.8\%$

Answer:

If the actual yield is $2.80$ g, the percent - yield is $79.8\%$ (you need to substitute the actual yield value you have to get the correct answer).