Question

PLEASE HELP

If 62.6 g of lead (II) chloride is produced, how many grams of lead (II) nitrate were reacted?

Answer 1

Answer: The mass of lead (II) nitrate required is 74.52 g

Step-by-step explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass.

The equation used is:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` ......(1)

Given mass of lead (II) chloride = 62.6 g

Molar mass of lead (II) chloride = 278.1 g/mol

Plugging values in equation 1:

`\text{Moles of lead (II) chloride}=(62.6g)/(278.1g/mol)=0.225 mol`

The chemical equation for the reaction of lead (II) chloride and sodium nitrate follows:

`Pb(NO_3)_2+2NaCl\rightarrow PbCl_2+2NaNO_3`

By the stoichiometry of the reaction:

1 mole of lead (II) chloride is produced from 1 mole of lead (II) nitrate

Then, 0.225 moles of lead (II) chloride will react with =
`(1)/(1)* 0.225=0.225mol` of lead(II) nitrate

Molar mass of lead (II) nitrate = 331.2 g/mol

Plugging values in equation 1:

`\text{Mass of lead (II) nitrate}=(0.225mol* 331.2g/mol)=74.52g`

Hence, the mass of lead (II) nitrate required is 74.52 g