Question

Starting with 0.657 g of lead(II) nitrate, a student collects 0.925 g of precipitate. If the calculated mass of precipitate is 0.914 g, what is the percent yield

Answer 1

Answer:

101.2%

Explanation:

Given:

Theoretical yield of the precipitate = 0.914 g

Actual yield of the precipitate = 0.925 g

Now, the percent yield is given as a ratio of actual yield by theoretical yield expressed as a percentage.

Framing in equation form, we have:

$\%\ yield=\frac{Actual\ yield}{Theoretical\ yield}\times 100$

Now, plug in 0.925 g for actual yield, 0.914 g for theoretical yield and solve for % yield. This gives,

$\%\ yield=\frac{0.925\ g}{0.914\ g}\times 100\\\\\%\ yield=1.012\times 100\\\\\%\ yield=101.2\%$

Therefore, the percent yield is 101.2%.