Answer
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:
$Ag^+(aq) + Cl^-(aq) -- \gt AgCl(s)$
