# Chapter 4 - Questions and Problems - Page 179: 4.83

The necessary $NaCl$ mass is equal to 0.165 g. Net ionic equation: $Ag^+(aq) + Cl^-(aq) -- \gt AgCl(s)$

#### Work Step by Step

1. Write the equation for that reaction: $NaCl(aq) + AgNO_3(aq) -- \gt AgCl(s) + NaNO_3(aq)$ It is already balanced. 2. Calculate the amount of $NaCl$ that is necessary. $250$ $mL$ $\times \frac{1L}{1000 mL} \times \frac{0.0113mol(AgNO_3)}{1L} = 0.002825 mol(AgNO_3)$ According to the balanced equation, the ratio of the reactants is 1 to 1. Therefore: Amount of $NaCl$ = $0.002825 mol$ 3. Calculate the molar mass for NaCl: Molar Mass ($NaCl$): 22.99* 1 + 35.45* 1 = 58.44g/mol 4. Find the mass in grams: $0.002285mol \times \frac{58.44}{1mol} = 0.165$ g. 5. Write the net ionic equation. $NaCl(aq) + AgNO_3(aq) -- \gt AgCl(s) + NaNO_3(aq)$ For the completely dissociated compounds, write their ions: $Na^+(aq) + Cl^-(aq) + Ag^+(aq) + NO_3^-(aq) -- \gt AgCl(s) + Na^+(aq) + NO_3^-(aq)$ Remove the spectators ions: $Ag^+(aq) + Cl^-(aq) -- \gt AgCl(s)$

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