Sodium hydroxide (NaOH\text{NaOH}NaOH) and sodium peroxide (Na2O2\text{Na}_2\text{O}_2Na2O2) have been used in closed environments like submarines to remove the carbon dioxide that crew members breathe out.
The equations for the reactions with carbon dioxide are:
2NaOH+CO2→Na2CO3+H2O 2\text{NaOH} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} 2NaOH+CO2→Na2CO3+H2O 2Na2O2+2CO2→2Na2CO3+O2 2\text{Na}_2\text{O}_2 + 2\text{CO}_2 \rightarrow 2\text{Na}_2\text{CO}_3 + \text{O}_2 2Na2O2+2CO2→2Na2CO3+O2Explain, with reference to these equations, two advantages of using sodium peroxide, rather than sodium hydroxide, to remove carbon dioxide from the air in a submarine.
Calculate the mass of sodium hydroxide needed to react with 150 g150\text{ g}150 g of carbon dioxide. [Mr of NaOH=40][M_r \text{ of NaOH} = 40][Mr of NaOH=40]
Calculate the volume of carbon dioxide, at room temperature and pressure (rtp), removed by 120 g120\text{ g}120 g of sodium peroxide. [Mr of Na2O2=78][M_r \text{ of Na}_2\text{O}_2 = 78][Mr of Na2O2=78] Assume that one mole of gas has a volume of 24 000 cm324\,000\text{ cm}^324000 cm3 at rtp.
80 exam-style questions on Edexcel IGCSE Chemistry Chemical formulae, equations and calculations. Each one has a worked solution and a mark scheme showing where the marks go.