A mixture contained CaCO3 and MgCo3. A sample of this mixture weighing 7.85g was reacted with excess HCl. The reactions are--

CaCO3 + 2HCl --> CaCl2 + H2O + CO2

MgCO3 + 2HCl --> MgCl2 + H2O + CO2

Is the sample reacted completely and produced 1.94L of CO2, at 25 degrees C, 785mmhg, whre were the percentages of CaCO3 and MgCO3 in the mixture?

So far, I got the moles of CO2 by using PV=nRT, which is .0817moles of CO2,

I don't kno what to do next.. can anyone PLEASE help!

Assume x grams of CaCO3, so 7.85-x grams of MgCO3.

Now, you can figure the moles of each (in terms of X), and then the moles of CO2 (in terms of x). Yes, it is messy algebra, but you add those moles and set equal to .0817, and you can solve for x. Double check each line of algebra.

Ca Co3(s) 2HCl (aq) Ca Cl2(aq)+H2o(e)+Co2(g)

To determine the percentages of CaCO3 and MgCO3 in the mixture, you need to find the moles of CaCO3 and MgCO3 that reacted in the reaction separately.

First, let's calculate the number of moles of CO2 produced. Given the volume (V) of CO2 gas, the temperature (T), and pressure (P), we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure of CO2 gas = 785 mmHg
V = volume of CO2 gas = 1.94 L
n = unknown moles of CO2
R = ideal gas constant = 0.0821 L.atm/(mol.K)
T = temperature = 25°C + 273.15 K = 298.15 K

Rearranging the equation, we get:

n = PV / RT

Substituting the values:

n = (785 mmHg * 1.94 L) / (0.0821 L.atm/(mol.K) * 298.15 K)
n = 0.0817 moles of CO2

We have already determined this value in your question.

Since the reactions for CaCO3 and MgCO3 state that 1 mole of each compound reacts to produce 1 mole of CO2, we know that the moles of CaCO3 and MgCO3 are equal to the moles of CO2.

Next, let's calculate the amount of CaCO3 and MgCO3 in the mixture.

The molar mass of CaCO3 is:
Ca = 40.08 g/mol (calcium)
C = 12.01 g/mol (carbon)
O = 16.00 g/mol (oxygen) x 3 atoms = 48.00 g/mol

Total = 100.09 g/mol

Using the molar mass of MgCO3:
Mg = 24.31 g/mol (magnesium)
C = 12.01 g/mol (carbon)
O = 16.00 g/mol (oxygen) x 3 atoms = 48.00 g/mol

Total = 84.32 g/mol

Since the moles of both CaCO3 and MgCO3 are equal to 0.0817, we can calculate their masses:

Mass of CaCO3 = 0.0817 moles * 100.09 g/mol = 8.17 grams (rounded to two decimal places)

Mass of MgCO3 = 0.0817 moles * 84.32 g/mol = 6.89 grams (rounded to two decimal places)

Finally, we can calculate the percentages:

Percentage of CaCO3 = (mass of CaCO3 / mass of mixture) * 100
Percentage of CaCO3 = (8.17 g / 7.85 g) * 100 ≈ 104.02%

Percentage of MgCO3 = (mass of MgCO3 / mass of mixture) * 100
Percentage of MgCO3 = (6.89 g / 7.85 g) * 100 ≈ 87.90%

Therefore, the percentages of CaCO3 and MgCO3 in the mixture are approximately 104.02% and 87.90%, respectively.

Note: The percentages you calculated are more than 100% because the initial assumption of complete reaction might not hold true in this calculation.