Lead(II) nitrate decomposes into lead(II) oxide, nitrogen dioxide, and oxygen according to the following balanced equation:
2Pb(NO3)2→2PbO+4NO2+O2 2\text{Pb}(\text{NO}_3)_2 \rightarrow 2\text{PbO} + 4\text{NO}_2 + \text{O}_2 2Pb(NO3)2→2PbO+4NO2+O2How many moles of products are formed if 12 mol of lead(II) nitrate decomposes?
12 mol12\text{ mol}12 mol of PbO\text{PbO}PbO, 12 mol12\text{ mol}12 mol of NO2\text{NO}_2NO2, 12 mol12\text{ mol}12 mol of O2\text{O}_2O2
12 mol12\text{ mol}12 mol of PbO\text{PbO}PbO, 24 mol24\text{ mol}24 mol of NO2\text{NO}_2NO2, 6 mol6\text{ mol}6 mol of O2\text{O}_2O2
24 mol24\text{ mol}24 mol of PbO\text{PbO}PbO, 12 mol12\text{ mol}12 mol of NO2\text{NO}_2NO2, 6 mol6\text{ mol}6 mol of O2\text{O}_2O2
24 mol24\text{ mol}24 mol of PbO\text{PbO}PbO, 24 mol24\text{ mol}24 mol of NO2\text{NO}_2NO2, 12 mol12\text{ mol}12 mol of O2\text{O}_2O2
91 exam-style questions on OCR GCSE Chemistry 5.1 Monitoring chemical reactions, covering 5.1.1 The technique of titration, 5.1.2 Theoretical mass, percentage yield and atom economy, 5.1.3 Concentration of a solution in mol/dm3, 5.1.4 Concentration calculations and titration results, 5.1.5 Molar gas volume and volumes of gases in reactions, and 5.1.6 Choosing a reaction pathway. Each one has a worked solution and a mark scheme showing where the marks go.