Question

(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.

Answer 1

(a) Molar mass of sodium aluminium fluoride = 3(23) + 27 + 6(19) = 210 g
Mass of sodium in sodium aluminium fluoride = 69 g
$\mathrm{Percentage}\mathrm{of}\mathrm{Na}=\frac{\mathrm{Mass}\mathrm{of}\mathrm{Na}\mathrm{in}\mathrm{compound}}{\mathrm{Molecular}\mathrm{mass}\mathrm{of}{\mathrm{Na}}_3{\mathrm{AlF}}_6}\times 100=\frac{69}{210}\times 100=32.85\approx 33\%$

(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