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)=(m_(HCl))/(n_(HCl)) \\ n_(HCl)=(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)=(m_(HCl))/(M_(HCl))=(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} (x)/(0.320891\ldots mol)=(1)/(2) \\ x=(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.