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2 years ago

A chemist performs the following reaction:

Chemistry

1 answer:

tester [92]2 years ago

4 0

Answer:

9.17 g

Explanation:

1) Calculate mols of PCl5

PCl5 (aq) + 4H2O (l) ⟶ H3PO4 (aq) + 5HCl (aq)

Mass of PCl5: 26.5g

Molar Mass of PCl5 208.24g/mol

Mol of PCl5 = Mass PCl5 /Molar mass PCl5 = 26.5g / (208.24g/mol) = 0.127257011 mol

2) Calculate mols of water needed to react

Mols of H2O per 1 Mols of PCl5 = 4 (because 4 water is needed to Phosphorus pentachloride).

Mols of H2O = Mol of PCl5 * 4 = 0.127257011 * 4 = 0.509028044 mols

3) Calculate Grams of water needed to react

Mols of H2O = 0.509028044 mols

Molar Mass of H2O = 18.015 g/mol

Mass of H2O = Mols of H2O * Molar Mass of H2O

= 0.509028044 mols* 18.015 g/mol = 9.17014021 g

sf = 3

9.17 g

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$\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm g\cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of $\rm AgCl$ formula units in $45.00\; \rm g$ of this compound:

$\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}$ .

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In other words, if ${\rm HCl}\, (aq)$ (the other reactant) is in excess, it would take exactly $1\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $1\; \rm mol \!$ of ${\rm AgCl}\, (s)\!$ formula units.

Hence, it would take $0.313987\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $0.313987\; \rm mol\!$ of ${\rm AgCl}\, (s)\!$ formula units.

Calculate the volume of the ${\rm AgNO_3} \, (aq)\!$ solution given that the concentration of the solution is $0.124\; \rm mol \cdot L^{-1}$ :

$\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}$ .

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of ${\rm AgNO_3} \, (aq)\!$ .)

In other words, approximately $2.53\; \rm L$ of that ${\rm AgNO_3} \, (aq)\!$ solution would be required.

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