The two nitrates undergo thermal decomposition on heating, as given by the equations below. $2\text{Pb(NO}_3)_2\text{(s)} \rightarrow 2\text{PbO(s)} + 4\text{NO}_2\text{(g)} + \text{O}_2\text{(g)}$ $2\text{NH}_4\text{NO}_3\text{(s)} \rightarrow 2\text{N}_2\text{(g)} + \text{O}_2\text{(g)} + 4\text{H}_2\text{O(l)}$ A separate sample of one mole of each nitrate is heated. The gas formed in each case is then passed through $\text{NaOH(aq)}$. After that, the volume of any gas that does not react with $\text{NaOH(aq)}$ is collected and measured. Which nitrate undergoes the larger percentage decrease in mass and gives the larger volume of gas collected?
- A$\text{NH}_4\text{NO}_3$; $\text{NH}_4\text{NO}_3$
- B$\text{NH}_4\text{NO}_3$; $\text{Pb(NO}_3)_2$
- C$\text{Pb(NO}_3)_2$; $\text{NH}_4\text{NO}_3$
- D$\text{Pb(NO}_3)_2$; $\text{Pb(NO}_3)_2$