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!

NEED HELP will NOT get you help. Name it Chemistry or whatever it is.

Sra

___C12H22O11 (s) + ___H2SO4 (aq) → ___C(s) + ___H2O (l) + ___H2SO4 (aq)

To determine the percentages of CaCO3 and MgCO3 in the mixture, we need to calculate the moles of each compound based on the moles of CO2 produced.

From the balanced chemical equations:

1 mole of CaCO3 produces 1 mole of CO2
1 mole of MgCO3 produces 1 mole of CO2

Given that you have 0.0817 moles of CO2 produced, we can assume that an equal number of moles of CaCO3 and MgCO3 were present because the reaction stoichiometry is 1:1 for both compounds.

Therefore, we have 0.0817 moles of both CaCO3 and MgCO3.

Now, to find the mass and percentage of each compound, we need to use the molar masses.

The molar mass of CaCO3 = (1 x molar mass of Ca) + (1 x molar mass of C) + (3 x molar mass of O)
= (1 x 40.08 g/mol) + (1 x 12.01 g/mol) + (3 x 16.00 g/mol)
= 100.09 g/mol

The mass of CaCO3 = moles of CaCO3 x molar mass of CaCO3
= 0.0817 moles x 100.09 g/mol
= 8.17 g

The molar mass of MgCO3 = (1 x molar mass of Mg) + (1 x molar mass of C) + (3 x molar mass of O)
= (1 x 24.31 g/mol) + (1 x 12.01 g/mol) + (3 x 16.00 g/mol)
= 84.32 g/mol

The mass of MgCO3 = moles of MgCO3 x molar mass of MgCO3
= 0.0817 moles x 84.32 g/mol
= 6.88 g

To calculate the percentages:

Percentage of CaCO3 = (mass of CaCO3 / mass of mixture) x 100%
= (8.17 g / 7.85 g) x 100%
= 104.05%

Percentage of MgCO3 = (mass of MgCO3 / mass of mixture) x 100%
= (6.88 g / 7.85 g) x 100%
= 87.74%

It seems that there might be an error in the calculations as the percentages are greater than 100%. Double-check the calculations to ensure the molar masses and masses are accurate.