3. From the equation: 3Cu + 8HNO > 3Cu(NOs)2+ 4H2o+ 2NO. Calculate(i)(ii)the mass of copper needed to react with 63 g of nitric acidthe volume of nitric oxide collected at the same time. [Cu-64, H-1, 0-16, N-1411

3. From the equation: 3Cu + 8HNO > 3Cu(NOs)2+ 4H2o+ 2NO. Calculate(i)(

1.	3. From the equation: 3Cu + 8HNO > 3Cu(NOs)2+ 4H2o+ 2NO. Calculate(i)(ii)the mass of copper needed to react with 63 g of nitric acidthe volume of nitric oxide collected at the same time. [Cu-64, H-1, 0-16, N-1411
Answer» Atomic weight of Cu= 63.5Molecular weight of HNO3= 1+14+3X16=633Cu + 8HNO3 → 3Cu(NO3)2 + 4H2O + 2NO3X63.5g =190.5g 8X63g=189g 2X22400ml=44800ml189g HNO3will react with 190.5g Cu 1g HNO3will react with (190.5/189) g Cu 63g HNO3will react with (190.5/189)X63 g Cu = 63.5g CuSo amount of copper needed is 63.5g.When 189g = 8mole HNO3will react to produce 2mole NO = 44800ml NO at STPWhen 189g HNO3react it will form 44800ml NO at STPWhen 1g HNO3react it will form 44800/189 ml NO at STPWhen 63g HNO3react it will form (44800/189)X63 ml NO at STP = 14933ml at STP = 14.933 L at STPSo at STP 14.933L NO will be collected.