Question

Lead (ii) carbonate decomposes to give lead (ii) oxide and carbon dioxide: pbco3 (s) ? pbo (s) + co2 (g) how many grams of lead (ii) oxide will be produced by the decomposition of 2.50 g of lead (ii) carbonate?

Answer 1

Answer : The mass of lead (ii) oxide produced by the decomposition of the reaction will be 2.09 grams.

Explanation : Given,

Mass of
`PbCO_3` = 2.50 g

Molar mass of
`PbCO_3` = 267.21 g/mole

Molar mass of
`PbO` = 223.2 g/mole

The given balanced chemical reaction is:

`PbCO_3(s)\rightarrow PbO(s)+CO_2(g)`

First we have to calculate the moles of
`PbCO_3`.

`\text{ Moles of }PbCO_3=\frac{\text{ Mass of }PbCO_3}{\text{ Molar mass of }PbCO_3}=(2.50g)/(267.21g/mole)=0.00936moles`

Now we have to calculate the moles of
`PbO`

From the reaction, we conclude that

As, 1 mole of
`PbCO_3` react to give 1 mole of
`PbO`

So, 0.0936 moles of
`PbCO_3` react to give 0.00936 moles of
`PbO`

Now we have to calculate the mass of
`PbO`

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

`\text{ Mass of }PbO=(0.00936moles)* (223.2g/mole)=2.09g`

Therefore, the mass of lead (ii) oxide produced by the decomposition of the reaction will be 2.09 grams.