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3 years ago

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Consider the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g). If 0.02 moles of PCl5, 0.04 moles of PCl3, and 0.08 moles of Cl2 are combined

in a 0.5 L flask, what is the reaction quotient (Q) before the reaction ? g

1 answer:

Furkat [3]3 years ago

4 0

Answer:

The reaction quotient (Q) before the reaction is 0.32

Explanation:

Being the reaction:

aA + bB ⇔ cC + dD

$Q=\frac{[C]^{c} *[D]^{d} }{[A]^{a}*[B]^{b} }$

where Q is the so-called reaction quotient and the concentrations expressed in it are not those of the equilibrium but those of the different reagents and products at a certain instant of the reaction.

The concentration will be calculated by:

$Concentration=\frac{number of moles of solute}{Volume}$

You know the reaction:

PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g).

So:

$Q=\frac{[PCl_{3} ] *[Cl_{2} ] }{[PCl_{5} ]}$

The concentrations are:

[PCl₃]= $\frac{0.04 moles}{0.5 L} =0.08 \frac{moles}{L}$

[Cl₂]= $\frac{0.08 moles}{0.5 L} =0.16 \frac{moles}{L}$

[PCl₅]= $\frac{0.02 moles}{0.5 L} =0.04 \frac{moles}{L}$

Replacing:

$Q=\frac{0.08*0.16}{0.04}$

Solving:

Q= 0.32

<u><em>The reaction quotient (Q) before the reaction is 0.32</em></u>

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Calcium Carbonate reacts with dilute hydrochloric acid. The equation for the reaction is shown. CaCo3 + 2Hcl = Cacl2 + H2O + Co2

Elena-2011 [213]

Answer:

Approximately $0.224\;\rm L$ , assuming that this reaction took place under standard temperature and pressure, and that $\rm CO_2$ behaves like an ideal gas. Also assume that the reaction went to completion.

Explanation:

The first step is to find out: which species is the limiting reactant?

Assume that $\rm CaCO_3$ is the limiting reactant. How many moles of $\rm CO_2$ would be produced?

Look up the relative atomic mass of $\rm Ca$ , $\rm C$ , and $\rm O$ on a modern periodic table:

$\rm Ca$ : $40.078$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass of $\rm CaCO_3$ :

$\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of formula units in $1\; \rm g$ of $\rm CaCO_3$ using its formula mass:

$\begin{aligned}& n(\mathrm{CaCO_3})\\&= \frac{m(\mathrm{CaCO_3})}{M(\mathrm{CaCO_3})} \\ &= \frac{1\; \rm g}{100.086\; \rm g \cdot mol^{-1}} \approx 1.00\times 10^{-2}\; \rm mol\end{aligned}$ .

In the balanced chemical equation, the ratio between the coefficient of $\rm CaCO_3$ and that of $\rm CO_2$ is $\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1$ .

In other words, for each mole of $\rm CaCO_3$ formula units consumed, one mole of $\rm CO_2$ would be produced.

If $\rm CaCO_3$ is indeed the limiting reactant, all that approximately $1.00\times 10^{-2}\; \rm mol$ of $\rm CaCO_3\!$ formula would be consumed. That would produce approximately $1.00\times 10^{-2}\; \rm mol\!$ of $\rm CO_2$ .

On the other hand, assume that $\rm HCl$ is the limiting reactant.

Convert the volume of $\rm HCl$ to $\rm dm^{3}$ (so as to match the unit of concentration.)

$\begin{aligned}&V(\mathrm{HCl})\\ &= 50\; \rm cm^{3} \\ &= 50\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{10^{3}\; \rm cm^{3}} \\ &= 5.00\times 10^{-2}\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $5.00\times 10^{-2}\; \rm dm^{3}$ of this $\rm 0.05\; \rm mol \cdot dm^{-3}$

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Therefore, if $\rm HCl$ is the limiting reactant, that $2.50 \times 10^{-3}\; \rm mol$ of $\rm HCl\!$ molecules would produce only one-half as many (that is, $1.25\times 10^{-3}\; \rm mol$ ) of $\rm CO_2$ molecules.

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In reality, no more than $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2$ molecules would be produced. The reason is that all $\rm CaCO_3$ would have been consumed before $\rm HCl$ was.

After finding the limiting reactant, approximate the volume of the $\rm CO_2\!$ produced.

Assume that this reaction took place under standard temperature and pressure (STP.) Under STP, the volume of one mole of ideal gas molecules would be approximately $22.4\; \rm L$ .

If $\rm CO_2$ behaves like an ideal gas, the volume of that $\rm 1.00\times 10^{-3}\; \rm mol$ of $\rm CO_2\!$ molecules would be approximately $\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L$ .

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