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(A) The combustion reaction is: $	ext{Organic compound} \xrightarrow{\Delta, CuO} CO_2 + H_2O$$1.$ Calculate the moles of $CO_2$ produced:$n_{CO_2} =

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$12$

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(A) The combustion reaction is: $	ext{Organic compound} \xrightarrow{\Delta, CuO} CO_2 + H_2O$$1.$ Calculate the moles of $CO_2$ produced:$n_{CO_2} = \frac{220 	imes 10^{-3} \ g}{44 \ g \ mol^{-1}} = 5 	imes 10^{-3} \ mol$$2.$ Calculate the mass of carbon in $CO_2$:Since $1 \ mol$ of $CO_2$ contains $1 \ mol$ of $C$,the moles of $C$ are $5 	imes 10^{-3} \ mol$.$m_{C} = 5 	imes 10^{-3} \ mol 	imes 12 \ g \ mol^{-1} = 60 	imes 10^{-3} \ g = 60 \ mg$$3.$ Calculate the percentage of carbon:$\% \ C = \frac{	ext{mass of } C}{	ext{mass of compound}} 	imes 100 = \frac{60 \ mg}{500 \ mg} 	imes 100 = 12 \%$

An organic compound weighing $500 \ mg$ produced $220 \ mg$ of $CO_2$ on complete combustion. The percentage composition of carbon in the compound is $..........\%$. $(nearest \ integer)$ $(Given \ molar \ mass \ in \ g \ mol^{-1} \ of \ C: 12, O: 16)$

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