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(A) Moles of $CO_2 = \frac{8.8 \ g}{44 \ g/mol} = 0.2 \ mol$. $1 \ mol$ of $CO_2$ contains $12 \ g$ of $C$. So,$0.2 \ mol$ of $CO_2$ contains $0.2 	i

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Law of Conservation of Mass

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(A) Moles of $CO_2 = \frac{8.8 \ g}{44 \ g/mol} = 0.2 \ mol$. $1 \ mol$ of $CO_2$ contains $12 \ g$ of $C$. So,$0.2 \ mol$ of $CO_2$ contains $0.2 	imes 12 = 2.4 \ g$ of $C$. Moles of $H_2O = \frac{5.4 \ g}{18 \ g/mol} = 0.3 \ mol$. $1 \ mol$ of $H_2O$ contains $2 \ g$ of $H$. So,$0.3 \ mol$ of $H_2O$ contains $0.3 	imes 2 = 0.6 \ g$ of $H$. Total mass of $C$ and $H$ in the hydrocarbon $= 2.4 \ g + 0.6 \ g = 3.0 \ g$. Since the mass of reactants $(3.0 \ g)$ equals the mass of products $(2.4 \ g \ C + 0.6 \ g \ H = 3.0 \ g)$,this follows the Law of Conservation of Mass.

When $3 \ g$ of a hydrocarbon is burnt in the presence of sufficient oxygen,it produces $8.8 \ g$ of $CO_2$ and $5.4 \ g$ of $H_2O$. This process is an example of which law?

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