X g of Ag is dissolved in HNO_3 and this sol | vedclass

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(B) The chemical reaction is: $Ag + HNO_3 \rightarrow AgNO_3 + NO + H_2O$ and $AgNO_3 + NaCl \rightarrow AgCl(s) + NaNO_3$.According to the law of con

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$2.16$

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(B) The chemical reaction is: $Ag + HNO_3 ightarrow AgNO_3 + NO + H_2O$ and $AgNO_3 + NaCl ightarrow AgCl(s) + NaNO_3$.According to the law of conservation of mass (stoichiometry),the moles of $Ag$ equal the moles of $AgCl$.Molar mass of $Ag = 108 \ g/mol$.Molar mass of $AgCl = 108 + 35.5 = 143.5 \ g/mol$.Moles of $AgCl = \frac{2.87 \ g}{143.5 \ g/mol} = 0.02 \ mol$.Since $1 \ mol$ of $Ag$ produces $1 \ mol$ of $AgCl$,moles of $Ag = 0.02 \ mol$.Mass of $Ag (X) = 0.02 \ mol 	imes 108 \ g/mol = 2.16 \ g$.

$X \ g$ of $Ag$ is dissolved in $HNO_3$ and this solution is treated with excess $NaCl$. If $2.87 \ g$ of $AgCl$ precipitate is obtained,then the value of $X$ is ............. $g$.

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