... nitrate was heated. The volume of oxygen formed was 0.16dm3 measured at r.t.p. The impurities did not decompose. Calculate the percentage of lead(II) nitrate in the sample. 2Pb(NO3)2 → 2PbO + 4NO2 + O2 Number of moles of O2 formed = 0.0067mol Number of moles of Pb(NO3)2 in... show more
Update: PS: This is the marking scheme for this question :
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if the final answer is between 86–89% award all 4
if the final answer is between 66–67% award 3 marks (Mr of 32 must have been used)
for all other answers marks can be awarded using the mark scheme as below and applying
ecf if necessary
number of moles of O2 formed = 0.16/24 = 0.0067/0.00667 or 1/150
number of moles of Pb(NO3)2 in the sample = 0.0133/0.013 or 1/75
mass of one mole of Pb(NO3)2 = 331 g
mass of lead(II) nitrate in the sample = 4.4(1) g
percentage of lead(II) nitrate in sample = 88.3% (allow 88–89) [4]
mark ecf in this question but not to simple integers
if mass of lead(II) nitrate 5.00 only marks 1 and 2 available
If divides by 32 (not 24) only last 3 marks can score consequentially