Question

How many grams of lead(II) sulfate (303 g/mol) are needed to react with sodium chromate (162 g/mol) in order to produce 0.162 kg of lead(II) chromate (323 g/mol)?

Answer 1

Answer : The mass of
`PbSO_4` needed are, 1.515 grams.

Explanation :

First we have to calculate the mole of
`PbCrO_4`.

`\text{Moles of }PbCrO_4=\frac{\text{Mass of }PbCrO_4}{\text{Molar mass of }PbCrO_4}=(0.162g)/(323g/mole)=0.005mole`

Now we have to calculate the moles of
`PbSO_4`.

The balanced chemical reaction will be,

produced from 1 mole of
`PbSO_4`

So, 0.005 mole of
`PbCrO_4` produced from 0.005 mole of
`PbSO_4`

Now we have to calculate the mass of
`PbSO_4`

`\text{Mass of }PbSO_4=\text{Moles of }PbSO_4* \text{Molar mass of }PbSO_4`

`\text{Mass of }PbSO_4=0.005mole* 303g/mole=1.515g`

Therefore, the mass of
`PbSO_4` needed are, 1.515 grams.