Answer to Question #132764 in General Chemistry for Abbey

Question

An impure sample of Pb(NO3)2 weighing 2.420 g was completely dissolved
in water. When all of the Pb+2 was precipitated from this solution with H2SO4 it produced 0.626 g of PbSO4(s). What was the percentage of Pb(NO3)2 in the sample?

Answer 1

The equation for the reaction is

$Pb(NO_3)_2+H_2SO_4\to PbSO_4+2HNO_3$

The ratio of $Pb(NO_3)_2$

$1$ atom of lead to $2$ atoms of nitrogen to $6$ atoms of Oxygen.

Simplified to $1Pb^{2+} to 2NO_3^{-}$

The molecular weight of $Pb(NO_3)_2=331.2g/mol$

Percentage yield$=$ $Actual yield\over theoretical yield$ $\times 100$

Maximum yield of $Pb(NO_3)_2$ $=$ $2.420g\times$ $207.2g\over (331.2\times 2)g$ $=0.756g$

Percentage yield $=$ $0.626g\over 0.756g$ $\times$ $100$

$=82.80\%$