Complete combustion of 0.66 g of a hydrocarb | vedclass

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(D) The molar mass of $CO_2 = 44 \ g/mol$. $44 \ g$ of $CO_2$ contains $12 \ g$ of carbon. $1.32 \ g$ of $CO_2$ contains $(1.32 	imes 12) / 44 = 0.36

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Conservation of Mass

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(D) The molar mass of $CO_2 = 44 \ g/mol$. $44 \ g$ of $CO_2$ contains $12 \ g$ of carbon. $1.32 \ g$ of $CO_2$ contains $(1.32 	imes 12) / 44 = 0.36 \ g$ of carbon. The molar mass of $H_2O = 18 \ g/mol$. $18 \ g$ of $H_2O$ contains $2 \ g$ of hydrogen. $2.7 \ g$ of $H_2O$ contains $(2.7 	imes 2) / 18 = 0.30 \ g$ of hydrogen. Total mass of carbon and hydrogen after combustion $= 0.36 \ g + 0.30 \ g = 0.66 \ g$. Since the total mass of the elements before and after the reaction remains the same,it follows the Law of Conservation of Mass.

List-$A$	List-$B$
$(1)$ Law of Multiple Proportions	$(A)$ Dalton
$(2)$ Law of Conservation of Mass	$(B)$ Joseph Proust
$(3)$ Law of Definite Proportions	$(C)$ Lavoisier
	$(D)$ Hofmann

Complete combustion of $0.66 \ g$ of a hydrocarbon gives $1.32 \ g$ of $CO_2$ and $2.7 \ g$ of water. The results of the given data follow the law of ...

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$(3)$ Law of Definite Proportions $(C)$ Lavoisier

$(D)$ Hofmann

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