42.6 g Cu are combined with 84.0 g of HNO3 according to the reaction:
3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O.
Which reagent is limiting and how many grams of Cu(NO3)2 are produced?

Question

42.6 g Cu are combined with 84.0 g of HNO3 according to the reaction: 
	3 Cu + 8 HNO3 → 3 Cu(NO3)2 + 2 NO + 4 H2O. 
Which reagent is limiting and how many grams of Cu(NO3)2 are produced?

anonymous · Answer

HNO3 limiting reagent and 93.76 g are produced

anonymous · Answer

Thanks how did you get that?

anonymous · Answer

wait one sec

anonymous · Answer

42.6gCU x 1 mole CU / 63.55gCU = .67 mole CU available 

84.0gHNO3 x 1mole HNO3 / 63.02gHNO3 = 1.33 mole HNO3 available

.67 mol CU x 8 mole HNO3(needed) / 3 mole CU =1.78 mole HNO3 needed and only 1.33 

mole available therefore HNO3 is Limiting.

84g HNO3 x 1 mole HNO3 / 63.02 HNO3 x 3 moles Cu(NO3)2 / 8 mole HNO3 x 187.57 g 

Cu(NO3)2 / 1 mole CU(NO3)2 =93.755 g of Cu(NO3)2 produced 

spaced it al out so it's easier to read

anonymous · Answer

Wow this process is very long...gotta study more! Thanks!

chmvijay · Answer

@nwynter now study more OK!!!!

anonymous · Answer

bleh is the equivalent of math and chem combined good luck