200cm3 each of 0.1M solutions of lead (II) trioxonitrate(V) and hydrochloric acid were mixed. Assuming that lead (II) chloride is completely insoluble, calculate the mass of lead (II) chloride that will be precipitated.
[Pb = 207, Cl= 35.5, N = 14, O = 16]
-
2.78g
- 5.56g
- 8.34g
- 11.12g
The correct answer is: A
Explanation
HCl + Pb(NO3)2 ===> 2HNO3 + PbCl2
2moles of HCl ===> 1mole of PbCl2
n = CV/1000 = (100 x 0.2)/(1000) = 0.02 mol
n (HCl) = 0.02 mol
2 moles of HCl ===> 1 mole of PbCl2
0.02 mole of HCl ===> x mole of PbCl2
2x = 0.02
x = 0.01 mol
n = m/M
m = 0.01 x (207 + (35.5 x 2)) = 0.01 x 278 = 2.78 g note; HCl is used for the calculation instead of lead(II) nitrate because it is the limiting reactant meaning it will be used up first before lead(11) nitrate
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