What mass of $lead chloride$ $"lead chloride"$ would result if a $21.5 \cdot m L$ $21.5mL$ volume of $lead nitrate$ $"lead nitrate"$ at $0.0806 \cdot m o l \cdot L^{- 1}$ $0.0806molL^-1$ concentration were mixed with a $18.00 \cdot m L$ $18.00mL$ volume of $potassium chloride$ $"potassium chloride"$ at $0.683 \cdot m o l \cdot L^{- 1}$ $0.683molL^-1$ concentration?

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What mass of $lead chloride$ $"lead chloride"$ would result if a $21.5 \cdot m L$ $21.5mL$ volume of $lead nitrate$ $"lead nitrate"$ at $0.0806 \cdot m o l \cdot L^{- 1}$ $0.0806molL^-1$ concentration were mixed with a $18.00 \cdot m L$ $18.00mL$ volume of $potassium chloride$ $"potassium chloride"$ at $0.683 \cdot m o l \cdot L^{- 1}$ $0.683molL^-1$ concentration?

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Answer 1 · 2017-06-20T03:08:58

Under a $2 \cdot g$ $2*g$ mass................

Explanation:

We need (i) a stoichiometric equation.....

$P b {(N O_{3})}_{2} (a q) + 2 K C l (a q) \to P b C l_{2} (s) ⏐ ↓ + 2 K N O_{3} (a q)$ $Pb(NO_3)_2(aq) + 2KCl(aq) rarr PbCl_2(s)darr+2KNO_3(aq)$

Lead chloride is one of the few water-insoluble halides. It crashes out of water solution with alacrity.

And (ii) we need equivalent quantities of each reagent......

$Moles of lead nitrate = 21.5 \times 10^{- 3} \cdot L \times 0.806 \cdot m o l \cdot L^{- 1} = 0.0173 \cdot m o l$ $"Moles of lead nitrate"=21.5xx10^-3*Lxx0.806*mol*L^-1=0.0173*mol$

$Moles of KCl = 18.00 \times 10^{- 3} \cdot L \times 0.683 \cdot m o l \cdot L^{- 1} = 0.0123 \cdot m o l$ $"Moles of KCl"=18.00xx10^-3*Lxx0.683*mol*L^-1=0.0123*mol$ .

And thus (clearly) chloride anion is the limiting reagent. Stoichiometry predicts that half an equiv of plumbous chloride will precipitate.

$Mass of$ $"Mass of"$ $P b C l_{2}$ $PbCl_2$ $=$ $=$ $\frac{0.0123 \cdot m o l}{2} \times 278.10 \cdot g \cdot m o l^{- 1}$ $(0.0123*mol)/2xx278.10*g*mol^-1$

$= 1.71 \cdot g$ $=1.71*g$

It is unusual that the chloride ion is in stoichiometric deficiency in that you would think we want to salt out the lead content......

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