0.43g limestone and 98cm3 of carbon dioxide are produced on reaction with excess hydrochloric acid. How much calcium carbonate is present?
You have to know pressure, temp to do this.
Standard temp and pressure
I don't get it. Limestone IS CaCO3. The reaction of limestone with HCl is as follows. You need to rephrase because limestone is not produced by reaction with HCl.
CaCO3 + 2HCl ==> CO2 + H2O + CaCl2
To determine how much calcium carbonate (CaCO3) is present, you need to use stoichiometry, which relates the number of moles of one substance to another in a balanced chemical equation.
Here's how you can calculate it step by step:
1. Write the balanced chemical equation for the reaction between hydrochloric acid (HCl) and calcium carbonate (CaCO3):
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O
You can see from the equation that one mole of CaCO3 produces one mole of CO2.
2. Convert the given mass of limestone (0.43g) to moles of CaCO3. To do this, you need to know the molar mass of CaCO3, which is the sum of the atomic masses of calcium (40.08 g/mol), carbon (12.01 g/mol), and oxygen (16.00 g/mol).
Molar mass of CaCO3 = 40.08 g/mol + 12.01 g/mol + (16.00 g/mol × 3) = 100.09 g/mol
Moles of CaCO3 = Mass (g) / Molar mass (g/mol) = 0.43 g / 100.09 g/mol ≈ 0.0043 mol
3. Since the reaction is with excess hydrochloric acid, all the moles of CaCO3 will react to produce the same number of moles of CO2.
Therefore, the moles of CO2 produced would be equal to the moles of CaCO3, which is 0.0043 mol.
4. Now, you need to convert the moles of CO2 to volume using the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L∙atm/(mol∙K)), and T is the temperature.
However, to use the ideal gas law, you need to know the pressure, temperature, and volume. In this case, you know the volume of CO2 produced (98 cm^3). But you need to convert it to liters by dividing by 1000:
Volume (L) = 98 cm^3 ÷ 1000 = 0.098 L
5. Rearranging the ideal gas law equation to solve for n (number of moles), you get:
n = PV / RT
Plugging in the known values:
n = (0.098 L) × (1 atm) / [(0.0821 L∙atm/(mol∙K)) × (298 K)] ≈ 0.00395 mol
6. Since the number of moles of CO2 produced is equal to the number of moles of CaCO3, the mass of CaCO3 can be calculated using its molar mass:
Mass of CaCO3 = Moles of CaCO3 × Molar mass of CaCO3
= 0.00395 mol × 100.09 g/mol
≈ 0.395 g
Therefore, approximately 0.395 grams of calcium carbonate is present.