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* 331.2g=662.4g` of lead nitrate produces
`4* 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=
`(662.4)/(184)* {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* 331.2g=662.4g` of lead nitrate produces 32 g of oxygen.

41.4 g of lead nitrate produces =
`(32)/(662.4)* {41.4}=2g` of oxygen.