What mass of calcium carbonate reaction with hydrochloric acid yields 44L of the gas at 27 ° C and 1520 mm Hg?
To find the mass of calcium carbonate that reacts with hydrochloric acid, we need to use the ideal gas law equation and stoichiometry.
Step 1: Write the balanced chemical equation for the reaction between calcium carbonate (CaCO3) and hydrochloric acid (HCl):
CaCO3 + 2HCl → CaCl2 + H2O + CO2
Step 2: Identify the gas involved in the reaction, which is carbon dioxide (CO2).
Step 3: Use the ideal gas law equation to calculate the number of moles of CO2:
PV = nRT
Where:
P = Pressure (1520 mm Hg)
V = Volume (44 L)
n = Number of moles
R = Ideal gas constant (0.0821 L·atm/mol·K)
T = Temperature in Kelvin (27 °C + 273.15 = 300.15 K)
Plugging in the values:
(1520 mm Hg) * (1 atm / 760 mm Hg) = 2 atm
(44 L) = V
(0.0821 L·atm/mol·K) = R
(300.15 K) = T
2 atm * (44 L) = n * (0.0821 L·atm/mol·K) * (300.15 K)
Solving for the number of moles (n):
n ≈ 3.69 moles
Step 4: Use the balanced chemical equation from Step 1 to determine the stoichiometric ratio between CaCO3 and CO2.
1 mole of CaCO3 produces 1 mole of CO2.
Since 1 mole of CaCO3 produces 1 mole of CO2, the number of moles of CaCO3 required will also be 3.69 moles.
Step 5: Calculate the molar mass of CaCO3. This can be found by adding up the atomic masses of each element in the compound:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 oxygen atoms in CO3)
Molar mass of CaCO3 = 40.08 + 12.01 + (16.00 * 3) ≈ 100.09 g/mol
Step 6: Calculate the mass of CaCO3 using the number of moles and molar mass:
Mass = moles * molar mass
Mass = (3.69 moles) * (100.09 g/mol) ≈ 369.20 g
Therefore, approximately 369.20 grams of calcium carbonate would react with hydrochloric acid to yield 44 L of carbon dioxide gas at 27°C and 1520 mm Hg.