A mixture of CaCO3 and MgCO3 weighing 7.85 g was reacted with excess HCL. The reactions are CaCO3+2HCL=CaCl2+H2O+CO2 and MgCO3+2HCl=MgCl2+H2O+CO2. The sample reacted completely to produce 1.94 L CO2 at 25 degrees C and 785 torr. Calculate the percentage of CaCO3 and MgCO3 in the original sample.

Two equations in two unknowns; solve simultaneously.

Let X = mass CaCO3
Let Y = mass MgCO3
======================
X + Y = 7.85 grams.
X(moles mass CO2/molar mass CaCO3) + Y(molar mass CO2/molar mass MgCO3) = grams CO2. [Note: YOu will need to use PV = nRT to calculate moles CO2, then convert that to grams CO2.]
Solve for X and Y then apply
%CaCO3 = (mass CaCO3/mass sample)*100 = ??
%MgCO3 = (mass MgCO3/mass sample)*100 = ??
Post your work if you get stuck.

Thank you. But could you explain how you got that equation for solving for x and y? I need to prove my answer.

.819 mol CO2= 30604 g CO2

X= 3.604 g CO2/100.09 g/molCaCO3= .036 mol CaCO3
Y=3.604 g CO2/84.32 g/mol MgCO3= .043 mol MgCO3

I'm stuck. I don't think I did it right.

To calculate the percentage of CaCO3 and MgCO3 in the original sample, we need to determine the number of moles of CO2 produced from each carbonate.

First, let's calculate the number of moles of CO2 produced using the ideal gas law:

PV = nRT

Where:
P = pressure (785 torr)
V = volume (1.94 L)
n = number of moles of CO2
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25 degrees Celsius = 25 + 273.15 = 298.15 K)

Rearranging the equation to solve for n:
n = PV / RT

Now let's calculate the number of moles of CO2 produced:

n = (785 torr * 1.94 L) / (0.0821 L·atm/(mol·K) * 298.15 K)

n ≈ 64.23 mol

Since the balanced equation shows that 1 mole of CaCO3 produces 1 mole of CO2, and 1 mole of MgCO3 produces 1 mole of CO2, we can determine the number of moles of CaCO3 and MgCO3 from the moles of CO2 produced.

Let's assume the mass of CaCO3 is x grams and the mass of MgCO3 is y grams:

1. Calculate the number of moles of CaCO3:
n(CaCO3) = x / molar mass(CaCO3)

Where the molar mass of CaCO3 is:
molar mass(CaCO3) = atomic mass(Ca) + atomic mass(C) + 3 * atomic mass(O)

Using atomic masses from the periodic table (rounded to two decimal places):
molar mass(CaCO3) = 40.08 g/mol + 12.01 g/mol + 3 * 16.00 g/mol ≈ 100.09 g/mol

Therefore:
n(CaCO3) = x / 100.09 g/mol

2. Calculate the number of moles of MgCO3:
n(MgCO3) = y / molar mass(MgCO3)

Where the molar mass of MgCO3 is:
molar mass(MgCO3) = atomic mass(Mg) + atomic mass(C) + 3 * atomic mass(O)

Using atomic masses from the periodic table (rounded to two decimal places):
molar mass(MgCO3) = 24.31 g/mol + 12.01 g/mol + 3 * 16.00 g/mol ≈ 84.31 g/mol

Therefore:
n(MgCO3) = y / 84.31 g/mol

Since the balanced equation shows that 1 mole of CaCO3 produces 1 mole of CO2, and 1 mole of MgCO3 produces 1 mole of CO2, we can set up the following equation:

n(CaCO3) + n(MgCO3) = 64.23 mol

Substituting the expressions for n(CaCO3) and n(MgCO3):

x / 100.09 g/mol + y / 84.31 g/mol = 64.23 mol

Simplifying the equation:

(x * 84.31) + (y * 100.09) = 64.23 * 84.31 * 100.09

This equation represents a linear system of two equations in two variables. Solving this system will give us the values of x and y.

The solution to this equation will provide the masses of CaCO3 (x) and MgCO3 (y) in the original sample. To calculate the percentage of each component, we can use the formula:

Percentage of CaCO3 = (mass of CaCO3 / total mass of the original sample) * 100%
Percentage of MgCO3 = (mass of MgCO3 / total mass of the original sample) * 100%