Element A reacts with oxygen to form a compou | vedclass

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(B) Given: Mass of $A = 0.359 \ g$. Mass of $A_2O_3 = 0.559 \ g$. Mass of oxygen = $0.559 - 0.359 = 0.2 \ g$. In $A_2O_3$,the molar ratio of $A$ to $O

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$43.08$

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(B) Given: Mass of $A = 0.359 \ g$. Mass of $A_2O_3 = 0.559 \ g$. Mass of oxygen = $0.559 - 0.359 = 0.2 \ g$. In $A_2O_3$,the molar ratio of $A$ to $O$ is $2:3$. Let the atomic mass of $A$ be $x$. Moles of $A = \frac{0.359}{x}$. Moles of $O = \frac{0.2}{16} = 0.0125$. According to the stoichiometry: $\frac{n_A}{n_O} = \frac{2}{3}$. $\frac{0.359 / x}{0.0125} = \frac{2}{3}$. $0.359 / x = 0.0125 	imes \frac{2}{3} = 0.008333$. $x = \frac{0.359}{0.008333} \approx 43.08 \ g/mol$.

Element $A$ reacts with oxygen to form a compound $A_2O_3$. If $0.359 \ g$ of $A$ reacts to form $0.559 \ g$ of the compound,the atomic mass of $A$ is:

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