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.
Related Chemistry A Level answers
All answers ▸Describe the Structure and Bonding of Benzene
Suggest why Phenol is more reactive than Benzene in Electrophillic Substitution
Part a) Draw the mechanism of the attack of CN- on CH3COCH3 in the presence of HCN Part b) Explain why the product of this reaction does not rotate the plane of plane polarised light
Explain the polarity of both CH3CH2Br and CBr3CBr3.