2 g of mixture of $CO\ and\ {CO}_{2}$ on reaction with excess $I_{2}O_{5}\ produced\ 2.54\ g\ of\ I_{2}.\$What would be the mass % of ${CO}_{2}$ in the original mixture?

Question

Dr. Abubacker · Accepted Answer

(b)
$5CO + I_{2}O_{5} ightarrow 5CO_{2} + I_{2}$
1 mole of $I_{2} \equiv$1 moles of $I_{2}O_{5} \equiv$5 moles of CO
Hence, mole of CO=$5 	imes \frac{2.54}{254} = 0.05$
Mass of CO=0.05$	imes 28 = 1.4$g
Mass of $CO_{2} = 2 - 1.4 = 0.6$ g
Mass % of $CO_{2} = \frac{0.6 	imes 100}{2} = 30$

Q. 2 g of mixture of $CO\ and\ {CO}_{2}$ on reaction with excess $I_{2}O_{5}\ produced\ 2.54\ g\ of\ I_{2}.\$What would be the mass % of ${CO}_{2}$ in the original mixture?

Explanation

Reviewed by Dr. Abubacker

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