Question

3. To form the precipitate PbCl2, 2.88 g of NaCl and 7.21 g of Pb(NO3)2 are mixed in solution. How much precipitate is formed? How much of which reactant is in excess?

Answer 1

$\begin{aligned} 2NaCl + Pb(NO_3)_2 \rightarrow PbCl_2 + 2NaNO_3 \end{aligned}$

$\begin{aligned} \textsf{From, number of moles (n) }&= \dfrac{\textsf{mass}}{\textsf{molar mass}}\\ \\ \textsf{Number of moles of NaCl } &= \dfrac{2.88}{58.5}\\ &= 0.049 moles\\ \\ \textsf{Number of moles of }Pb(NO_3)_2 &= \dfrac{7.21}{331} \\&= 0.0218 moles \end{aligned}$

1 mole of$Pb(NO_3)_2$ reacts with 2 moles of $NaCl$

0.0218 moles of$Pb(NO_3)_2$ reacts with 0.0436 moles of$NaCl$

$\therefore NaCl$ is in excess, since the amount of $NaCl$ in the reaction is 0.049 moles.

$\begin{aligned} &1 \textsf{ mole of }Pb(NO_3)_2 =\textsf{ 2 moles of }PbCl_2\\ &\therefore 0.0218 \textsf{ moles of }Pb(NO_3)_2 =\textsf{ 0.0218} \textsf{ moles of }PbCl_2\\ \\ &\textsf{mass of }PbCl_2 = \textsf{ no. of moles of }PbCl_2 ×\textsf{ molar mass of }PbCl_2\\ &\textsf{mass of }PbCl_2 \textsf{ produced }= \ 0.0218 × (207 + 2(35))\\ \end{aligned}$$\begin{aligned} \qquad \qquad \qquad \qquad &= 0.0218 ×271\\ \qquad \qquad \qquad \qquad &=5.91g \end{aligned}$

$\therefore 5.91$g of precipitate is formed