Cu(s) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l) 1.000 g copper wire is rea

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? 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? 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?

Explanation / Answer

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

1.00 g of Cu = 1.00/ 63.54 mol of Cu = 0.01574 mol of Cu

10 mL of 16 M HNO3 = 10 x 16/ 1000 mol of HNO3 = 0.16 mol of HNO3

So Cu is the limiting reagent in this case,

therefore % yield = (1.000 - 0.953) x 100/ 1.000 = 4.70 %

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