Question

if 0.953 g of copper was actually recovered at the end of the entire experiment, what would be the percent yield?

Answer 1

Here is the full question:

Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

1.000 g copper wire is reacted with 10 mL of concentrated nitric acid (16M). If 0.953 g of copper was actually recovered at the end of the entire experiment, what would be the percent yield?

Answer:

95.30 %

Explanation:

equation for the reaction is given as:

Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)

1.00 g of Cu = $\frac{1.00}{63.54 mol of Cu}$

= 0.01574 mol of Cu

10 mL of 16 M HNO3 = $10*\frac{16}{1000 mole of HNO3}$

= 0.16 mol of HNO3

However, from both variables; we can arrive at a conclusion that Cu is the limiting reagent.

therefore % yield = = 4.70 %

Percentage yield = $\frac{actual yield}{theorectical yield} *100%$

Percentage yield = $\frac{0.953}{ 1.000}* 100$%

Percentage yield = 0.953 × 100%

Percentage yield = 95.30 %