Please help me and show me step by step. assume that the reactions are being performed at pressure of 1.2 atm and a temperature of 298K
Calcium carbonate decomposes at a high temperature to form carbon dioxide and calcium oxide. How many grams of calcium carbonate will be nedded to form 3.45 L of carbon dioxide?
1. Write the equation and balance it.
CaCO3 ==> CO2 + CaO
2. Use PV = nRT to calculate n = number of moles CO2. You know P, V, T, R.
3. Using the coefficients in the balanced equation, convert moles CO2 to moles CaCO3.
4. Now convert moles CaCOP3 to grams. g = moles x molar mass.
To determine the number of grams of calcium carbonate needed to form 3.45 L of carbon dioxide, we need to use the ideal gas law equation as follows:
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 K)
First, let's convert the given temperature from degrees Celsius to Kelvin:
T = 298K
Next, let's rearrange the ideal gas law equation to solve for the number of moles:
n = PV / RT
Substituting the known values:
P = 1.2 atm
V = 3.45 L
T = 298 K
R = 0.0821 L.atm/mol.K
n = (1.2 atm * 3.45 L) / (0.0821 L.atm/mol.K * 298 K)
Now, let's calculate the value of n by dividing the numerator by the denominator:
n = 0.164 mol
The balanced chemical equation for the decomposition of calcium carbonate is:
CaCO3 -> CO2 + CaO
From the balanced equation, we can see that 1 mole of calcium carbonate decomposes to give 1 mole of carbon dioxide. Therefore, the number of moles of calcium carbonate required to produce 0.164 mol of carbon dioxide is also 0.164 mol.
To convert moles of calcium carbonate to grams, we need to use its molar mass. The molar mass of calcium carbonate (CaCO3) is:
Ca: 40.08 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 atoms)
Adding up these masses, we get:
(40.08 g/mol) + (12.01 g/mol) + (16.00 g/mol * 3) = 100.09 g/mol
Now, let's calculate the mass of calcium carbonate required:
Mass = number of moles * molar mass
Mass = 0.164 mol * 100.09 g/mol
Finally, solve for Mass by multiplying the number of moles by the molar mass:
Mass = 16.405 g
Therefore, approximately 16.405 grams of calcium carbonate are needed to produce 3.45 L of carbon dioxide at a pressure of 1.2 atm and a temperature of 298K.