# Question f83f5

Jul 12, 2016

#### Answer:

Here's how you can do that.

#### Explanation:

Your starting point here will be the balanced chemical equation that describes this reaction

${\text{CaCO"_ (3(s)) + color(red)(2)"HCl"_ ((aq)) -> "CaCl"_ (2(aq)) + "CO"_ (2(g)) uarr + "H"_ 2"O}}_{\left(l\right)}$

Notice that the reaction consumes $\textcolor{red}{2}$ moles of hydrochloric acid and produces $1$ mole of carbon dioxide for every mole of calcium carbonate that takes part in the reaction.

Now, use calcium carbonate's molar mass to calculate how many moles you have in your $\text{1.00 g}$ sample

1.00 color(red)(cancel(color(black)("g"))) * "1 mole CaCO"_3/(100.09color(red)(cancel(color(black)("g")))) = "0.009992 moles CaCO"_3

Use the molarity and volume of the hydrochloric acid solution to calculate how many moles you have in the sample

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, you have

n_("HCl") = "2.00 mol" color(red)(cancel(color(black)("dm"^(-3)))) * overbrace(20.0 * 10^(-3)color(red)(cancel(color(black)("dm"^3))))^(color(blue)("volume in dm"^3))

$\textcolor{w h i t e}{a} = \text{0.0400 moles HCl}$

As you can see, you have more than enough moles of hydrochloric acid present in the solution to allow for all the moles of calcium carbonate to react, i.e. calcium carbonate acts as a limiting reagent.

You can say that because $0.009992$ moles of calcium carbonate would only require

0.009992 color(red)(cancel(color(black)("moles CaCO"_3))) * (color(red)(2)color(white)(a)"moles HCl")/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = 0.01998 ~~ "0.0200 moles HCl"

So, you know that the reaction consumes $0.009992$ moles of calcium carbonate, which means that it produces

0.009992color(red)(cancel(color(black)("moles CaCO"_3))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole CaCO"_3)))) = "0.009992 moles CO"_2

Now, more often than not, STP conditions are given to students as a pressure of $\text{1 atm}$ and a temperature of ${0}^{\circ} \text{C}$. Under these conditions for pressure and temperature, one mole of any ideal gas occupies $\text{22.4 L}$ $\to$ this is known as the molar volume of a gas at STP.

It's worth mentioning that those conditions for pressure and temperature correspond to the old definition of STP. The current definition has a pressure of $\text{100 kPa}$ and a temperature of ${0}^{\circ} \text{C}$.

Under these conditions for pressure and temperature, one mole of any ideal gas occupies $\text{22.7 L}$.

I'll do the calculation using the current definition, and thus the value of $\text{22.7 L/mol}$ at STP.

0.009992 color(red)(cancel(color(black)("moles CO"_2))) * overbrace("22.7 L"/(1color(red)(cancel(color(black)("mole CO"_2)))))^(color(blue)("molar volume of a gas at STP")) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.227 L")color(white)(a/a)|)))#

The answer is rounded to three sig figs.

SIDE NOTE If you were given the old STP definition, and thus the old value for the molar volume of a gas at STP, simply redo the last calculation using $\text{22.4 L}$ instead of $\text{22.7 L}$.

Jul 13, 2016

#### Answer:

$V = 0.224 \setminus L$

#### Explanation:

First, write a balanced chemical equation for the reaction.

$C a C {O}_{3} + 2 H C l \to C {O}_{2} + {H}_{2} O + C a C {l}_{2}$

Then find the number of moles of each of the two reactants used.

• Moles of $C a C {O}_{3}$ used.

$1.00 \setminus g \setminus C a C {O}_{3} \times \frac{1 \setminus m o l . C a C {O}_{3}}{100.09 \setminus g \setminus C a C {O}_{3}}$

$1.00 \setminus \cancel{g \setminus C a C {O}_{3}} \times \frac{1 \setminus m o l . C a C {O}_{3}}{100.09 \setminus \cancel{g \setminus C a C {O}_{3}}}$

$0.00999 \setminus m o l . C a C {O}_{3}$

• Moles of $H C l$ used

$n = {C}_{M} \times V$

$2.00 \setminus \frac{m o l .}{{\mathrm{dm}}^{3}} \times 0.0200 \setminus {\mathrm{dm}}^{3}$

$2.00 \setminus \frac{m o l .}{\cancel{\left({\mathrm{dm}}^{3}\right)}} \times 0.0200 \setminus \cancel{{\mathrm{dm}}^{3}}$

$0.0400 \setminus m o l . H C l$

Note that one of the two reactants is in excess. For this reason we find the number of moles of the $C {O}_{2}$ that could be produced by each of the two reactants. The smallest value is the one to be considered. It represents the maximum amount that could be produced from the limiting reactant.

$0.00999 \setminus m o l . C a C {O}_{3} \times \frac{1 \setminus m o l . C {O}_{2}}{1 \setminus m o l . C a C {O}_{3}}$

$0.00999 \setminus \cancel{m o l . C a C {O}_{3}} \times \frac{1 \setminus m o l . C {O}_{2}}{\cancel{1 \setminus m o l . C a C {O}_{3}}}$

$0.00999 \setminus m o l . C {O}_{2}$

$0.0400 \setminus m o l . H C l \times \frac{1 \setminus m o l . C {O}_{2}}{2 \setminus m o l . H C l}$

$0.0400 \setminus \cancel{m o l . H C l} \times \frac{1 \setminus m o l . C {O}_{2}}{2 \setminus \cancel{m o l . H C l}}$

$0.0200 \setminus m o l . C {O}_{2}$

So the maximum amount of $C {O}_{2}$.that could be produced is $0.00999 \setminus m o l .$

Under S.T.P conditions, we can use the following formula to calculate the volume of the $C {O}_{2}$ produced.

${n}_{C {O}_{2}} = \frac{V}{V} _ M$

${n}_{C {O}_{2}} \text{ }$ Is the number of moles.

$V \text{ }$ is the volume occupied by the gas.

${V}_{M} \text{ }$ is the molar volume. It is the volume occupied
$\text{ }$by one mol. of any gas under S.T.P conditions and it is
$\text{ }$equal to 22.4 L.

$V = {n}_{C {O}_{2}} \times {V}_{M}$

$V = 0.00999 \setminus m o l . \times 22.4 \setminus \frac{L}{m o l .}$

$V = 0.00999 \setminus \cancel{m o l .} \times 22.4 \setminus \frac{L}{\cancel{m o l .}}$

$V = 0.224 \setminus L$