Carbon dioxide can be made in the lab by the reaction of hydrochloric acid with calcium carbonate. How many milliliters of dry CO2 at 20.0 °C and 745 torr can be prepared from a mixture of 30.6 g of CaCO3 and 279 mL of 0.250 M HCl?
This is a limiting reagent (LR) problem.
CaCO3 + 2HCl ==> H2O + CO2 + CaCl2
mols CaCO3 = grams/molar mass
mols CO2 = mols CaCO3 (from the coefficients in the balanced equation).
mols HCl = M x L = ?
mols CO2 = 1/2 that (do the coefficient thing again.)
The values for mols CO2 produced probably will not agree which means one of them is not right. In LR problems the smaller value is ALWAYS the correct value and the reagent producing that value is the limiting reagent.
Using the smaller value for n, use PV = nRT and solve for V in liters and convert to mL.
NOTE!!!!. I highly suspect that the person making up this problem wants you to correct for the vapor pressure of H2O; however, the way the problem is stated that correction is not necessary. The way I have worked the problem is for dry CO2 and says nothing about how the CO2 is collected. IF it is collected over water AND has these same parameters, then a correction is necessary.
Use PV = nRT and the conditions lis
To determine the volume of dry carbon dioxide (CO2) that can be prepared in this reaction, we need to use the concept of stoichiometry. First, let's write and balance the chemical equation for the reaction:
CaCO3 + 2HCl → CO2 + H2O + CaCl2
Now, let's calculate the number of moles of CaCO3 and HCl:
Molar mass of CaCO3 = 40.08 g/mol (Ca) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 100.09 g/mol
Number of moles of CaCO3 = mass / molar mass = 30.6 g / 100.09 g/mol ≈ 0.305 moles
Next, we need to find the limiting reagent to determine which reactant is completely consumed. The ratio between HCl and CaCO3 in the balanced equation is 2:1. So, let's calculate the number of moles of HCl:
Molarity (M) = moles / volume (L)
Number of moles of HCl = Molarity × volume (L)
= 0.250 mol/L × 0.279 L
≈ 0.06975 moles
Since the molar ratio is 2:1, we can see that the HCl is the limiting reagent because we have fewer moles of HCl compared to CaCO3.
Now, let's determine the number of moles of CO2 produced. Since the molar ratio between CaCO3 and CO2 is 1:1, the number of moles of CO2 is also 0.305 moles.
Finally, we can use the ideal gas law to calculate the volume of CO2 at 20.0 °C and 745 torr. The ideal gas law equation is:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in L)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
Temperature (K) = 20.0 °C + 273.15 = 293.15 K
Now we can plug the values into the equation:
V = (nRT) / P
= (0.305 mol × 0.0821 L·atm/(mol·K) × 293.15 K) / 0.986 atm (since 1 atm = 0.986 torr)
≈ 7.21 L
Finally, we need to convert the volume from liters to milliliters:
Volume (mL) = 7.21 L × 1000 mL/L
≈ 7210 mL
Therefore, approximately 7210 milliliters of dry CO2 can be prepared at 20.0 °C and 745 torr.