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

101.2%

Step-by-step 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=(Actual\ yield)/(Theoretical\ yield)* 100`

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

`\%\ yield=(0.925\ g)/(0.914\ g)* 100\\\\\%\ yield=1.012* 100\\\\\%\ yield=101.2\%`

Therefore, the percent yield is 101.2%.