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natta225 [31]
3 years ago
14

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

1 g of calcium Carbonate is added to 50cm3 of 0.05mol/dm3 hydrochloric acid. Which volume of carbon dioxide is made in this reaction?
Chemistry
1 answer:
Elena-2011 [213]3 years ago
3 0

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}

\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}.

Notice that in the balanced chemical reaction, the ratio between the coefficient of \rm HCl and that of \rm CO_2 is \displaystyle \frac{n(\mathrm{CO_2})}{n(\mathrm{HCl})} = \frac{1}{2}.

In other words, each mole of \rm HCl molecules consumed would produce only 0.5\;\rm mol of \rm CO_2 molecules.

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.

If \rm CaCO_3 is the limiting reactant, \rm 1.00\times 10^{-3}\; \rm mol of \rm CO_2 molecules would be produced. However, if \rm HCl is the limiting reactant, 1.25\times 10^{-3}\; \rm mol of \rm CO_2\! molecules would be produced.

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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