Question

A mixture contained calcium carbonate and magnesium carbonate in unspecified proportions. A 7.85g sample of this mixture has reacted with excess hydrochloric acid, producing 1.94L of carbon dioxide at 25 degrees C and 785mmHg. What are the percentage of Calcium Carbonate and Magnesium Carbonate in the sample?

Answer 1

To determine the percentage of calcium carbonate and magnesium carbonate in the mixture, you need to calculate the moles of carbon dioxide produced from each carbonate and convert them to grams. Then, use the moles of CO2 to calculate the percentages of each carbonate in the sample.

Step-by-step explanation:

To determine the percentage of calcium carbonate and magnesium carbonate in the given sample, we need to calculate the moles of carbon dioxide produced from each carbonate and then convert them to grams. First, we calculate the moles of CO2 produced using the ideal gas law equation, PV = nRT. We can assume the volume is at STP (standard temperature and pressure) conditions, so we substitute the given values (1.94L of CO2, 25 degrees C, 785mmHg) and solve for n. Next, we use the balanced chemical equation to relate the moles of CO2 produced to the moles of calcium carbonate and magnesium carbonate in the sample.

Let's assume the sample contains x grams of calcium carbonate and y grams of magnesium carbonate. To find the percentage of each carbonate in the sample, we calculate the moles of CO2 produced from each carbonate and then divide by the total moles of CO2 produced. Finally, we multiply these values by 100 to get the percentages.

Answer 2

Answer: The percentage of calcium carbonate is 76.3% and magnesium carbonate is 23.7%.

Step-by-step explanation:

`CO_2` produced ca be calculated from ideal gas equation:

`PV=nRT`

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

`n=(PV)/(RT)=(1.03* 1.94L)/(0.0821 * 298)=0.08moles`

Mass of
`CO_2={\text{Given moles}}* {\text {Molar mass}}=0.08moles* 44g/mol=3.6g`

If the mass of calcium carbonate is x g , (7.85-x) g of magnesium carbonate is there.

`CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2`

As, 100 g of
`CaCO_3` react to give 44 g of
`CO_2` gas

So, x g of
`CaCO_3` react to give
`(44* x)/(100)g` of
`CO_2` gas

The balanced chemical reaction will be:

`MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2`

As, 84 g of
`MgCO_3` react to give 44 g of
`CO_2` gas

So,
`(7.85-x)g` of
`MgCO_3` react to give
`(44* (7.85-x))/(84)g` of
`CO_2` gas

`(44* x)/(100)+(44* (7.85-X))/(84)=3.6`

By solving the terms, we get the value of x

x = 5.99 g

The mass of
`CaCO_3` = x = 5.99 g

The mass of
`MgCO_3` = 7.85 - 5.99 = 1.86 g

Now we have to calculate the percentage of magnesium carbonate.

`\%\text{ of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Total mass}}* 100=(5.99g)/(7.85g)* 100=76.3\%`

`\%\text{ of }MgCO_3=\frac{\text{Mass of }MgCO_3}{\text{Total mass}}* 100=(1.86g)/(7.85g)* 100=23.7\%`

Therefore, the percentage of calcium carbonate is 76.3% and magnesium carbonate is 23.7%.