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

Sagot :

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

Explanation:

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

The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(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:

[tex]\text{Moles of lead (II) chloride}=\frac{62.6g}{278.1g/mol}=0.225 mol[/tex]

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

[tex]Pb(NO_3)_2+2NaCl\rightarrow PbCl_2+2NaNO_3[/tex]

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 = [tex]\frac{1}{1}\times 0.225=0.225mol[/tex] of lead(II) nitrate

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

Plugging values in equation 1:

[tex]\text{Mass of lead (II) nitrate}=(0.225mol\times 331.2g/mol)=74.52g[/tex]

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

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