Question

How many grams of lead (II) oxide will be produced by the decomposition of 2.50 g of lead (II) carbonate? *

Answer 1

The balanced equation for considered reaction can be written as follows

$PbCO_3 \xrightarrow{t \degree C} PbO + CO_2$

$\frac{m(PbCO_3)}{M(PbCO_3) \cdot n(PbCO_3)} = \frac{m(PbO)}{M(PbO) \cdot n(PbO)}$,

where $m$ - mass, g; $M$ - molar mass, g/mol; $n$ - number of molecules in the balanced equation.

$m(PbO) = \frac{m(PbCO_3) \cdot M(PbO) \cdot n(PbO)}{M(PbCO_3) \cdot n(PbCO_3)}$

$M(PbCO_3) = 1 \cdot 207.2 + 1 \cdot 12.011 + 3 \cdot 15.9994 \approx 267.21 \; g/mol$

$M(PbO) = 1 \cdot 207.2 1 \cdot 15.9994 \approx 223.21 \; g/mol$

$m(PbO) = \frac{2.50 \cdot 223.21 \cdot 1}{267.21 \cdot 1} \approx 2.09 \; g$

Answer: 2.09 g.