Question

CaCO3 + 2HCl → CaCl2 + CO2 + H2OHow many grams of CO2 are produced if 11.7 g of HCl react?

Answer 1

To answer this, we will need:

1: convert the mass of HCl to number of moles of HCl;

2: make the stoichimetry to see the corresponding number of moles of CO₂;

3: convert the number of moles of CO₂ to mass.

To convert the mass to number of moles, we will use:

$\begin{gathered} M_{HCl}=\frac{m_{HCl}}{n_{HCl}} \\ n_{HCl}=\frac{m_{HCl}}{M_{HCl}}_{} \end{gathered}$

The molar mass of HCl can be calculated using the molar masses of its atoms:

$\begin{gathered} M_{HCl}=1\cdot M_H+1\cdot M_{Cl} \\ M_{HCl}=(1\cdot1.00794+1\cdot35.453)g/mol \\ M_{HCl}=(1.00794+35.453)g/mol \\ M_{HCl}=36.46094g/mol \end{gathered}$

So:

$n_{HCl}=\frac{m_{HCl}}{M_{HCl}}=\frac{11.7g}{36.46094g/mol}=0.320891\ldots mol$

Now, the ratio of the number of moles of the components of a reaction has to obey the stoichiometry.

Accouding to the reaction equation, 2 moles of HCl produce 1 mol of CO₂.

Thus, if x is the number of moles of CO₂ produced:

x --- 1 mol CO₂

0.320891... mol HCl --- 2 mol HCl

$\begin{gathered} \frac{x}{0.320891\ldots mol}=\frac{1}{2} \\ x=\frac{1}{2}\cdot0.320891mol \\ x=0.160445\ldots mol \end{gathered}$

Now, we just need to convert the number of moles to mass:

$\begin{gathered} n_{CO_2}=x=0.160445\ldots mol \\ M_{CO_{2}}=\frac{m_{CO_2}}{n_{CO_{2}}} \\ m_{CO_2}=M_{CO_2}\cdot n_{CO_{2}} \end{gathered}$

The molar mass of CO₂ is:

$\begin{gathered} M_{CO_2}=1\cdot M_C+2\cdot M_O \\ M_{CO_2}=(1\cdot12.0107+2\cdot15.9994)g/mol \\ M_{CO_2}=(12.0107+31.9988)g/mol \\ M_{CO_2}=44.0095g/mol \end{gathered}$

So:

$m_{CO_2}=M_{CO_2}\cdot n_{CO_2}=44.0095g/mol\cdot0.160445\ldots mol=7.0611\ldots g\approx7.06g$

Thus, assuming all the HCl react, approximately 7.06 g of CO₂ will be produced.