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(a) Calculate the percentage of sodium in sodium aluminium fluoride (Na3AlF6) 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 ⟶ 2CO2

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

Mole Concept

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Answer

Molecular weight of sodium aluminium fluoride (Na3AlF6)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g

210 g of sodium aluminium fluoride contains 69 g of Na

∴ 100 g of sodium aluminium fluoride will contain = 69210\dfrac{69}{210} x 100 = 32.85% = 33%

Hence, percentage of sodium in sodium aluminium fluoride (Na3AlF6) is 33%

(b)

2CO+O2⟶2CO22 vol.:1 vol.⟶2 vol.\begin{matrix} 2\text{CO} & + & \text{O}2 & \longrightarrow & 2\text{CO}2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \\ \end{matrix}

1 mole of O2 has volume = 22400 ml

Volume of oxygen used by 2 × 22400 ml CO = 22400 ml

∴ Vol. of O2 used by 560 ml CO

= 224002×22400\dfrac{22400}{2 × 22400} x 560

= 280 ml

Volume of CO2 formed by 2 × 22400 ml CO = 2 x 22400 ml

∴ Vol. of CO2 formed by by 560 ml CO

= 560 ml

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