How many grams of lead oxide will be produced by the decomposition of 267.2g of lead carbonat?

Answer 1

First of all it is 'Lead Carbonate' ; note the 'e' on the end of 'carbonate'. 

For the thermal decomposition of lead carbonate , write up the BALANCED reaction equation. 

PbCO3 =ht=> PbO + CO2

The molar ratios are 1::1:1 

Next calculate the Mr of PbCO3 & PbO 

PbCO3

Pb x 1 = 207 x 1 = 207

C x 1 = 12 x 1 = 12 

Ox 3 = 16 x 3 = 48 

207 + 12 + 48 = 267 

PbO 

Pb x 1 = 207 x 1 = 207

O x 1 = 16 x 1 = 16

207 + 16 = 223 

Using the Equation 

moles = mass(g) / Mr 

mol(PbCO3) = 267.2/267 = 1.00075 moles (Equivalent to one molar ratio)

mol(PbO) = 1.00075 moles( Also equivalent to '1' molar ratio)

Hence 

1.00075 = mass(g) / 223 

mass(PbO) = 1.00075 x 223 = 223.167 g