Question

Time Remaining:03:55:13 1. Consider the following balanced equation: 3CuO(s)+ 2NH3(g)3H2O()+3Cu(s) + N2(g) If 80.9 moles of CuO(s) and 31.4 moles of NH3(g) are allowed to react, and the percent yield is 90.9%, how many moles of Cu(s) will actually be produced? 62.4 moles 56.7 moles 40.5 moles 42.8 moles 80.0 moles Use the slider to rate your confidence on this question. -1 100 Not at all confident Very confident Irc 55 44 R save subit Rut 101

Answer 1

Balanced chemical equation is:

3 CuO + 2 NH3 ---> 3 Cu + 4 H2O

3 mol of CuO reacts with 2 mol of NH3

for 80.9 mol of CuO, 53.93 mol of NH3 is required

But we have 31.4 mol of NH3

so, NH3 is limiting reagent

According to balanced equation

mol of Cu formed = (3/2)* moles of NH3

= (3/2)*31.4

= 47.1 mol

% yield = actual mol*100/theoretical mol

90.9= actual mol*100/47.1

actual mol = 42.8 moles

Answer: 42.8 moles