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=(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
`(6)/(3)* 1.27=2.54` mole of
`NaCl`

Now we have to calculate the mass of
`NaCl`

`\text{ Mass of }NaCl=\text{ Moles of }NaCl* \text{ Molar mass of }NaCl`

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

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