Question #853bc

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Question #853bc

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Answer 1 · 2014-12-11T12:28:15

The answer is $0.12 g$ $0.12g$ .

Essentially, what you are dealing with is a double replacement precipitation reaction in which two soluble ionic compounds (two soluble salts) react to form an insoluble precipitate.

Starting from the balanced chemical equation, we get

$N a C l_{(a q)} + A g N O_{3 (a q)} \to N a N O_{3 (a q)} + A g C l_{(s)}$ $NaCl_((aq)) + AgNO_(3(aq)) -> NaNO_(3(aq)) + AgCl_((s))$

The complete ionic equation, which better describes the reaction since ionic compounds dissociate into ions when dissolved in water, is

$A g_{(a q)}^{+} + N O_{3 (a q)}^{-} + N a_{(a q)}^{+} + C l_{(a q)}^{-} \to A g C l_{(s)} + N a_{(a q)}^{+} + N O_{3 (a q)}^{-}$ $Ag_((aq))^(+) + NO_(3(aq))^(-) + Na_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s)) + Na_((aq))^(+) + NO_(3(aq))^(-)$

Notice that $N a_{(a q)}^{+}$ $Na_((aq))^(+)$ and $N O_{3 (a q)}^{-}$ $NO_(3(aq))^(-)$ did not participate in the reaction since they can be found on both sides of the equation -> such ions are called spectator ions.

This gives us the net ionic equation, which shows us what ions participate in the reaction that forms the precipitate

$A g_{(a q)}^{+} + C l_{(a q)}^{-} \to A g C l_{(s)}$ $Ag_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s))$

Now, we know that we have a $1 : 1$ $1:1$ mole ratio for $N a C l$ $NaCl$ and $A g N O_{3}$ $AgNO_3$ ; the number of moles of $A g N O_{3}$ $AgNO_3$ can be determined from molarity, $C = \frac{n_{s o l u t e}}{V_{s o l u t i o n}}$ $C = n_(solute)/(V_(solution)$

$n_{s o l u t e} = C \cdot V_{s o l u t i o n} = \frac{0.100 m o l e s}{L} \cdot (20 \cdot 10^{- 3}) L$ $n_(solute) = C * V_(solution) = (0.100 mol es)/(L) * (20 * 10^(-3)) L$ = 0.0020 moles

This means that the number of moles of $N a C l$ $NaCl$ must be 0.0020 as well. Knowing $N a C l$ $NaCl$ 's molar mass - $58.5 \frac{g}{m o l}$ $58.5 g/(mol)$ - we get

$m_{N a C l} = 0.0020 m o l e s \cdot 58.5 \frac{g}{m o l} = 0.12 g$ $m_(NaCl) = 0.0020 mol es * 58.5 g/(mol) = 0.12g$

This is true because the number of moles of $N a C l$ $NaCl$ dissociate into moles of $N a^{+}$ $Na^+$ and moles of $C l^{-}$ $Cl^(-)$ , the same being true for $A g N O_{3}$ $AgNO_3$ ; therefore, the number of $N a C l$ $NaCl$ moles must be at least equat to the number of $A g N O_{3}$ $AgNO_3$ moles.

A more in depth analysis could be done using the concentrations of the compounds, the reaction's coefficient ( $Q_{s p}$ $Q_(sp)$ ), and $K_{s p}$ $K_(sp)$ .

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