Respuesta :
Answer: The percentage of calcium carbonate is 76.3% and magnesium carbonate is 23.7%.
Explanation:
[tex]CO_2[/tex] produced ca be calculated from ideal gas equation:
[tex]PV=nRT[/tex]
P= Pressure of the gas = 785 mmHg = 1.03 atm   (760 mmHg= 1 atm)
V= Volume of the gas = 1.94 L
T= Temperature of the gas = 25°C=(25+273)K= 298 K  (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
[tex]n=\frac{PV}{RT}=\frac{1.03\times 1.94L}{0.0821 \times 298}=0.08moles[/tex]
Mass of [tex]CO_2={\text{Given moles}}\times {\text {Molar mass}}=0.08moles\times 44g/mol=3.6g[/tex]
If the mass of calcium carbonate is x g , (7.85-x) g of magnesium carbonate is there.
[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2[/tex]
As, 100 g of [tex]CaCO_3[/tex] react to give 44 g of [tex]CO_2[/tex] gas
So, x g of [tex]CaCO_3[/tex] react to give [tex]\frac{44\times x}{100}g[/tex] of [tex]CO_2[/tex] gas
The balanced chemical reaction will be:
[tex]MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2[/tex]
As, 84 g of [tex]MgCO_3[/tex] react to give 44 g of [tex]CO_2[/tex] gas
So, [tex](7.85-x)g[/tex] of [tex]MgCO_3[/tex] react to give [tex]\frac{44\times (7.85-x)}{84}g[/tex] of [tex]CO_2[/tex] gas
[tex]\frac{44\times x}{100}+\frac{44\times (7.85-X)}{84}=3.6[/tex]
By solving the terms, we get the value of x
x = 5.99 g
The mass of [tex]CaCO_3[/tex] = x = 5.99 g
The mass of [tex]MgCO_3[/tex] = 7.85 - 5.99 = 1.86 g
Now we have to calculate the percentage of magnesium carbonate.
[tex]\%\text{ of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Total mass}}\times 100=\frac{5.99g}{7.85g}\times 100=76.3\%[/tex]
[tex]\%\text{ of }MgCO_3=\frac{\text{Mass of }MgCO_3}{\text{Total mass}}\times 100=\frac{1.86g}{7.85g}\times 100=23.7\%[/tex]
Therefore, the percentage of calcium carbonate is 76.3% and magnesium carbonate is 23.7%.
