Question

42.6g of Cu are combined with 48.0g of HNO3, according to the reaction Cu HNO3 Cu(NO32 NO H2 Which reactant is limiting and how many grams of Cu(NOs) are produced?

Answer 1

The second question is out of topic, so I'm going to answer thr first one only.

Given reaction: 3Cu + 8HNO Vertical-bar _3 rightarrow 3 cu(NO_3)_2 + 2NO + 4 H_2O Moles of cu provide = 42.6/63.5 = 0.671 Moles of HNO_3 provided = 48/63 = 0.762 Now, 3 moles of cu use 8 moles of HNO_3 0.67 moles of Cu needs = 8/3 times 0.67 = 1.79 moles of HNO_0 But, we only have 0.762 moles of HNO_3 Hence, HNO_3 is the limiting reagent Grams of Cu(NO_3)_2 produced by 0.762 moles HNO_3 2 moles HNO_3 gives: 3 moles cu(NO_2)_2 & 0.762 moles HNO_3 give = 3/8 times 0.762 = 0.29 moles Weight of 0.29 moles Cu(NO_3)_2: 0.29 times molecular weight = 0.29 times 187.5 = 54.375 g