Question

If 354.2g PbCl2 are reacted with excess Na3PO4, how many grams of NaCl will be produced

Answer 1

Answer: The mass of $NaCl$ produced is, 149 grams.

Explanation : Given,

Mass of $PbCl_2$ = 354.2 g

Molar mass of $PbCl_2$ = 278 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

First we have to calculate the moles of $PbCl_2$

$\text{Moles of }PbCl_2=\frac{\text{Given mass }PbCl_2}{\text{Molar mass }PbCl_2}$

$\text{Moles of }PbCl_2=\frac{354.2g}{278g/mol}=1.27mol$

Now we have to calculate the moles of $CaCl_2$

The balanced chemical equation is:

$3PbCl_2+2Na_3PO_4\rightarrow 6NaCl+Pb_3(PO_4)_2$

From the reaction, we conclude that

As, 3 moles of $PbCl_2$ react to give 6 mole of $NaCl$

So, 1.27 moles of $PbCl_2$ react to give $\frac{6}{3}\times 1.27=2.54$ mole of $NaCl$

Now we have to calculate the mass of $NaCl$

$\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl$

$\text{ Mass of }NaCl=(2.54moles)\times (58.5g/mole)=148.59g\approx 149g$

Therefore, the mass of $NaCl$ produced is, 149 grams.