# (i) What volume of 5.00*mol*L^-1 HCl(aq) is required to react with a 17.0*g mass of sodium hydroxide. (ii) what volume of sodium sulfide at 0.100*mol*L^-1, is required to precipitate a 33.0*mL volume of 0.100*mol*L^-1 of AgNO_3?

Jul 26, 2017

We need to find (i) write a stoichiometrically balanced equation, and (ii) we need to find the molar quantities of each reagent.

#### Explanation:

Our basic equation is......

$\text{Concentration"="Moles of solute"/"Volume of solution}$.

And thus by taking a product or a quotient, we can get the number of moles or the volume of solution.

$N a O H \left(s\right) + H C l \left(a q\right) \rightarrow N a C l \left(a q\right) + {H}_{2} O \left(l\right)$

And thus, for the first equation there is 1:1 stoichiometry. And moles of $N a O H$ are equivalent to moles of $H C l$.

$\text{Moles of NaOH} \equiv \frac{17.0 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1} = 0.425 \cdot m o l$

And by the give stoichiometry, we need ONE equiv of $H C l \left(a q\right)$.

We are given $5.00 \cdot m o l \cdot {L}^{-} 1$ $H C l \left(a q\right)$, and thus we solve the quotient......

$\frac{0.425 \cdot \cancel{m o l}}{5.00 \cdot \cancel{m o l \cdot {L}^{-} 1}} \times {10}^{3} \cdot m L \cdot \cancel{{L}^{-} 1} = 85.0 \cdot m L$.

Clear?

For the second scenario, you must simply KNOW that $A {g}_{2} S$ (like most sulfides save those of the alkali metals) is INSOLUBLE.

$2 A g N {O}_{3} \left(a q\right) + N {a}_{2} S \left(a q\right) \rightarrow A {g}_{2} S \left(s\right) \downarrow + 2 N a N {O}_{3} \left(a q\right)$

The net ionic equation is simply......

$2 A {g}^{+} + {S}^{2 -} \rightarrow A {g}_{2} S \left(s\right) \downarrow$

$\text{Moles of}$ $A g N {O}_{3} = 33.0 \times {10}^{-} 3 \cdot L \times 0.100 \cdot m o l \cdot {L}^{-} 1 =$
$3.30 \times {10}^{-} 3 \cdot m o l$

And thus we need half an equiv of $N {a}_{2} S$, i.e.

$\frac{\frac{1}{2} \times 3.3 \times {10}^{-} 3 \cdot m o l}{0.100 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1$

$\text{under 17 mL}$ of the sodium sulfide solution.........