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# (a) Calculate the percentage of sodium in sodium aluminium fluoride (Na3AIF6) correct to the nearest whole number. (F = 19; Na = 23; Al = 27) (b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows: 2CO + O2 $\stackrel{}{\to }$ 2CO2 Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

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Solution

## (a) Molar mass of sodium aluminium fluoride = 3(23) + 27 + 6(19) = 210 g Mass of sodium in sodium aluminium fluoride = 69 g $\mathrm{Percentage}\mathrm{of}\mathrm{Na}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Na}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{Na}}_{3}{\mathrm{AlF}}_{6}}×100=\frac{69}{210}×100=32.85\approx 33%$ (b) For two moles of CO, one mole of O2 is used. So, according to the Avogadro's law: Volume of oxygen used by 560 mL of CO = $\frac{560}{2}\mathrm{mL}=280\mathrm{ml}$ Also, number of moles or volume of CO used in reaction is equal to the number of moles or volume of CO2 formed. Volume of CO2 formed = 560 mL  Suggest Corrections  0      Similar questions  Explore more