$(i)$ $(i)$ What volume of $5.00 \cdot m o l \cdot L^{- 1}$ $5.00molL^-1$ $H C l (a q)$ $HCl(aq)$ is required to react with a $17.0 \cdot g$ $17.0g$ mass of sodium hydroxide. $(i i)$ $(ii)$ what volume of sodium sulfide at $0.100 \cdot m o l \cdot L^{- 1}$ $0.100molL^-1$ , is required to precipitate a $33.0 \cdot m L$ $33.0mL$ volume of $0.100 \cdot m o l \cdot L^{- 1}$ $0.100molL^-1$ of $A g N O_{3}$ $AgNO_3$ ?

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Answer 1 · 2017-07-26T07:43:31

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

Our basic equation is......

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

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

$N a O H (s) + H C l (a q) \to N a C l (a q) + H_{2} O (l)$ $NaOH(s) + HCl(aq) rarr NaCl(aq) + H_2O(l)$

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

$Moles of NaOH \equiv \frac{17.0 \cdot g}{40.0 \cdot g \cdot m o l^{- 1}} = 0.425 \cdot m o l$ $"Moles of NaOH"-=(17.0*g)/(40.0*g*mol^-1)=0.425*mol$

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

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

$(0.425*cancel(mol))/(5.00*cancel(mol*L^-1))xx10^3*mL*cancel(L^-1)=85.0*mL$ .

Clear?

For the second scenario, you must simply KNOW that $Ag_2S$ (like most sulfides save those of the alkali metals) is INSOLUBLE.

$2AgNO_3(aq) + Na_2S(aq) rarr Ag_2S(s)darr+2NaNO_3(aq)$

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

$2Ag^+ + S^(2-) rarr Ag_2S(s)darr$

$"Moles of"$ $AgNO_3=33.0xx10^-3*Lxx0.100*mol*L^-1=$
$3.30xx10^-3*mol$

And thus we need half an equiv of $Na_2S$ , i.e.

$(1/2xx3.3xx10^-3*mol)/(0.100*mol*L^-1)xx10^3*mL*L^-1$

$"under 17 mL"$ of the sodium sulfide solution.........