what is the percent yield of nitrogen monoxide when 57.4 g of copper reacts with 1.58 x 10^24 molecules of nitric acid to produce 15 g of nitrogen monoxide. Answer is 83%
6Cu + 8HNO3 > 2NO +4H2O + 6CU(NO3)2
Boiiiiiiii lol you on the wrong website for that question.
Kudos to anyone who can help you tho lol
In my chem class. whenever i dont know what to do just start with dimensional analysis. The limiting reactant in this reaction is the copper so the moles of the nitric acid is negligible. hoped that help
B. Paul George