When excess silver nitrate is added to 5.850 | vedclass

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Question

(D) The balanced chemical equation for the reaction is:$NaCl + AgNO_3 \rightarrow AgCl + NaNO_3$From the stoichiometry of the reaction:$1 \ mol$ of $N

Vedclass · Accepted Answer

$14.35$

Vedclass · Answer

(D) The balanced chemical equation for the reaction is:$NaCl + AgNO_3 ightarrow AgCl + NaNO_3$From the stoichiometry of the reaction:$1 \ mol$ of $NaCl$ produces $1 \ mol$ of $AgCl$.Molar mass of $NaCl = 23 + 35.5 = 58.5 \ g/mol$.Number of moles of $NaCl = \frac{5.85 \ g}{58.5 \ g/mol} = 0.1 \ mol$.Since $1 \ mol$ of $NaCl$ produces $1 \ mol$ of $AgCl$,$0.1 \ mol$ of $NaCl$ will produce $0.1 \ mol$ of $AgCl$.Molar mass of $AgCl = 107.8 + 35.5 = 143.3 \ g/mol$ (using $143.5 \ g/mol$ as per standard approximation).Mass of $AgCl = 0.1 \ mol 	imes 143.5 \ g/mol = 14.35 \ g$.

When excess silver nitrate is added to $5.850 \ g$ of sodium chloride,how many grams of silver chloride are produced?

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