An excess of Lead (II) oxide reacts with 175cm3 of 1.5 mol dm3 nitric acid. Calculate the maximum quantity of lead that can be obtained from this reaction.
Write the equation: PbO + HNO3 --> PbNO3 + H2O. Balance the equation: there are two moles of hydrogen (in water) on the right. To balance the hydrogens, two moles of hydrogen are needed on the left to HNO3. PbO + 2HNO3 --> PbNO3 + H2O. Now we have two moles of nitrogen and six moles of oxygen in 2HNO3 (on the left); we need to balance this by adding two moles to the right. PbO + 2HNO3 --> Pb (NO3)2 + H2O. Once we have written and balanced the equation, we can move to the next part of the question. Find the moles of HNO3: Volume of nitric oxide =175 x 10-3 Concentration of nitric oxide= 1.5 mol dm3. Moles of HNO3 = 175 x 10-3 x 1.5= 0.2625 moles. Find the moles of Pb(NO3) 2: as there were two moles of HNO3 (see balanced equation above), we can work out the moles of Pb(NO3)2 by dividing the moles of HNO3 by two, 0.2625/2=0.131 moles of Pb(NO3)2. Work out the Mr of Pb(NO3)2: 207.2+ (16 x 61) + (2 x 14) = 331.2g mol-1. Work out the mass of Pb(NO3) 2: mass= mole x Mr 0.131 x 331.2 = 43.5g The maximum quantity of lead that can be obtained from the reaction is 43.5 grams.
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