Answer :
Answer:
Approximately [tex]0.224\;\rm L[/tex], assuming that this reaction took place under standard temperature and pressure, and that [tex]\rm CO_2[/tex] 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 [tex]\rm CaCO_3[/tex] is the limiting reactant. How many moles of [tex]\rm CO_2[/tex] would be produced?
Look up the relative atomic mass of [tex]\rm Ca[/tex], [tex]\rm C[/tex], and [tex]\rm O[/tex] on a modern periodic table:
- [tex]\rm Ca[/tex]: [tex]40.078[/tex].
- [tex]\rm C[/tex]: [tex]12.011[/tex].
- [tex]\rm O[/tex]: [tex]15.999[/tex].
Calculate the formula mass of [tex]\rm CaCO_3[/tex]:
[tex]\begin{aligned} & M(\rm CaCO_3) \\ &= 40.078 + 12.011 + 3 \times 15.999 \\&= 100.086\; \rm g \cdot mol^{-1}\end{aligned}[/tex].
Calculate the number of moles of formula units in [tex]1\; \rm g[/tex] of [tex]\rm CaCO_3[/tex] using its formula mass:
[tex]\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}[/tex].
In the balanced chemical equation, the ratio between the coefficient of [tex]\rm CaCO_3[/tex] and that of [tex]\rm CO_2[/tex] is [tex]\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{CaCO_3})} = 1[/tex].
In other words, for each mole of [tex]\rm CaCO_3[/tex] formula units consumed, one mole of [tex]\rm CO_2[/tex] would be produced.
If [tex]\rm CaCO_3[/tex] is indeed the limiting reactant, all that approximately [tex]1.00\times 10^{-2}\; \rm mol[/tex] of [tex]\rm CaCO_3\![/tex] formula would be consumed. That would produce approximately [tex]1.00\times 10^{-2}\; \rm mol\![/tex] of [tex]\rm CO_2[/tex].
On the other hand, assume that [tex]\rm HCl[/tex] is the limiting reactant.
Convert the volume of [tex]\rm HCl[/tex] to [tex]\rm dm^{3}[/tex] (so as to match the unit of concentration.)
[tex]\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}[/tex].
Calculate the number of moles of [tex]\rm HCl[/tex] molecules in that [tex]5.00\times 10^{-2}\; \rm dm^{3}[/tex] of this [tex]\rm 0.05\; \rm mol \cdot dm^{-3}[/tex]
[tex]\begin{aligned}& n(\mathrm{HCl}) \\ &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.05\; \rm mol \cdot dm^{-3}\\ &\quad\quad \times 5.00\times 10^{-2}\;\rm dm^{3} \\ &= 2.50 \times 10^{-3}\; \rm mol\end{aligned}[/tex].
Notice that in the balanced chemical reaction, the ratio between the coefficient of [tex]\rm HCl[/tex] and that of [tex]\rm CO_2[/tex] is [tex]\displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}[/tex].
In other words, each mole of [tex]\rm HCl[/tex] molecules consumed would produce only [tex]0.5\;\rm mol[/tex] of [tex]\rm CO_2[/tex] molecules.
Therefore, if [tex]\rm HCl[/tex] is the limiting reactant, that [tex]2.50 \times 10^{-3}\; \rm mol[/tex] of [tex]\rm HCl\![/tex] molecules would produce only one-half as many (that is, [tex]1.25\times 10^{-3}\; \rm mol[/tex]) of [tex]\rm CO_2[/tex] molecules.
If [tex]\rm CaCO_3[/tex] is the limiting reactant, [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2[/tex] molecules would be produced. However, if [tex]\rm HCl[/tex] is the limiting reactant, [tex]1.25\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2\![/tex] molecules would be produced.
In reality, no more than [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2[/tex] molecules would be produced. The reason is that all [tex]\rm CaCO_3[/tex] would have been consumed before [tex]\rm HCl[/tex] was.
After finding the limiting reactant, approximate the volume of the [tex]\rm CO_2\![/tex] 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 [tex]22.4\; \rm L[/tex].
If [tex]\rm CO_2[/tex] behaves like an ideal gas, the volume of that [tex]\rm 1.00\times 10^{-3}\; \rm mol[/tex] of [tex]\rm CO_2\![/tex] molecules would be approximately [tex]\rm 1.00\times 10^{-3}\; \rm mol \times 22.4\; \rm L = 0.224\; \rm L[/tex].