(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 $\overset{}{\to}$ 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
$Percentage of Na = \frac{Mass of Na in compound}{Molecular mass of {Na}_{3} {AlF}_{6}} \times 100 = \frac{69}{210} \times 100 = 32.85 \approx 33 %$

(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 = $\frac{560}{2} mL = 280 ml$
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

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560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:

$2 C O + O_{2} \to 2 C O_{2}$

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:

$2 C O + O_{2} \to 2 C O_{2}$

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:

$2 C O + O_{2} \to 2 C O_{2}$

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

[N a_{3} A l F_{6}]

F = 19; N a = 23; A l = 27

Calculate the percentage of sodium in sodium aluminium fluoride (Na₃AlF₆) correct to the nearest whole number.

(F= 19,Na = 23,Al = 27).

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