Answer to Question #141837 in Chemistry for Autumn Murphy

Question

Calculate the mass of lead in 139g of Pb(NO3)2

Answer 1

According to the chemical formula of Pb(NO₃)₂, where 1 formula unit contains 1 atom of lead, the number of the moles of lead nitrate and the number of the moles of lead are equal:

$n(Pb(NO_3)_2) =n(Pb)$ .

The number of the moles of lead nitrate contained in 139 g is the ratio of its mass and its molar mass $M=331.21$ g/mol:

$n(Pb(NO_3)_2) = \frac{m}{M} = \frac{139 \text{g}}{331.21\text{g/mol}} = 0.4197$ mol.

Therefore, the number of the moles of lead is :

$n(Pb) = 0.4197$ mol.

The molar mass of lead is 207.20 g/mol. Thus, its mass contained in 139 g of lead nitrate is:

$m(Pb) = nM = 0.4197\text{mol}\cdot207.20\text{g/mol}=86.96$ , or 87 g.

Answer: 87 g of lead is contained in 139 g of Pb(NO₃)₂.