If 1g of copper reacts with 10mL of nitric acid according to the equation: Cu(s) + 4HNO3(aq) -->Cu(NO3)2(aq) + 2No2(g) +2H2O(l)What mass of copper would be produced? (if it were isolated)
Cu + 4 HNO3 → Cu(NO3)2 + 2 NO2 + 2 H2O Copper is not produced, so supposing the question to be: What mass of copper nitrate would be produced? (1 g Cu) / (63.5463 g Cu/mol) = 0.0157 mol Cu (0.0157 mol Cu) x (4/1) / (0.010 L) = 6.28 mol/L As long as the concentration of the nitric acid is greater than 6.3 M (which is reasonable since concentrated HNO3 is about 15.8 M) copper is the limiting reactant. (0.0157 mol Cu) x (1/1) x (187.5563 g Cu(NO3)2/mol) = 3 g Cu(NO3)2 (If the concentration of the acid is less than 6.28 M, it becomes the limiting reactant and it's concentration must be specified in order to calculate the product(s) obtained.)
zinc oxide + hydrochloric acid ---> zinc chloride + water ZnO + HCl ---> ZnCl2 +H2O zinc oxide + sulphuric acid ---> zinc sulphate + water ZnO + H2SO4 ---> ZnSO4 + H2O zinc oxide + nitric acid ---> ZnO + HNO3 ---> ZnN + H2O + O
