Question

Can anyone help me with this?

The equation below shows the decomposition of lead nitrate. How many grams of oxygen are produced when 23.0 g NO2 is formed? 2Pb(NO3)2(s) _ 2PbO(s) + 4NO2(g) + O2 (g

Answer 1

Molecular mass of NO2 is (14 + 2 x 32=) 46 therefore 1 mole of NO2 is 46g
and 28.75 is 28.75/46+0.625 moles
Which equates to 0.625 x 32= 20 g oxygen. (1mole of O2 is 2x16=32 grams)
But oxygen/NO2 ratio is 1:4 so 1/4x20=5g produced

Answer 2

2Pb(NO3)2(s) -> 2PbO(s) + 4NO2(g) + O2 (g)

1) molar masses of O2 and NO2

O2: 32 g/mol

NO2: 14 + 32 = 46 g/mol

2) Conversions

4 mol NO2 = 4mol * 46g/mol = 184 g
1 mol O2 = 32 g

3) Ratios

184 g NO2 / 32 g O2 = 23 g NO2 / x g O2 => x = 23 g NO2 * [ 32 g O2 / 184 g NO2] = 4 g O2.

Answer: 4 g.