Question

the equation below shows the decomposition of lead nitrate. how many grams of oxygen are produced when 11.5g NO2 is formed?

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

Answer 1

Answer: 2 g

Explanation: $2Pb(NO_3)_2(s)\rightarrow 2PbO(s)+4NO_2(g)+O_2(g)$

As can be seen from the balanced chemical equation, 2 moles of lead nitrate produce 4 moles of nitrogen dioxide.

$2\times 331.2g=662.4g$ of lead nitrate produces $4\times 46=184g$ of nitrogen dioxide.

184 g of nitrogen dioxide will be produced by 662.4 g of lead nitrate

So 11.5 g of nitrogen dioxide will be produced by=$\frac{662.4}{184}\times {11.5}=41.4 g$ of lead nitrate

As can be seen from the balanced chemical equation, 2 moles of lead nitrate produce 1 mole of oxygen.

$2\times 331.2g=662.4g$ of lead nitrate produces 32 g of oxygen.

41.4 g of lead nitrate produces =$\frac{32}{662.4}\times {41.4}=2g$ of oxygen.