Answer :
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 = [tex]\frac{1.00}{63.54 mol of Cu}[/tex]
= 0.01574 mol of Cu
10 mL of 16 M HNO3 = [tex]10*\frac{16}{1000 mole of HNO3}[/tex]
= 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 = [tex]\frac{actual yield}{theorectical yield} *100%[/tex]
Percentage yield = [tex]\frac{0.953}{ 1.000}* 100[/tex]%
Percentage yield = 0.953 × 100%
Percentage yield = 95.30 %