For the reaction below, Kp = 1.16 at 800.°C.
CaCO3(s) CaO(s) + CO2(g)
If a 21.7-g sample of CaCO3 is put into a 10.6 L container and heated to 800°C, what percentage by mass of the CaCO3 will react to reach equilibrium?
Please calculate
To find the percentage by mass of CaCO3 that will react to reach equilibrium, we need to first calculate the number of moles of CaCO3 in the 21.7 g sample. Then we'll use the balanced chemical equation to determine the moles of CaCO3 that will react to reach equilibrium.
Step 1: Calculate the number of moles of CaCO3
To determine the number of moles, we'll use the molar mass of CaCO3.
The molar mass of CaCO3 = molar mass of Ca + molar mass of C + 3 * (molar mass of O)
The molar mass of Ca = 40.08 g/mol
The molar mass of C = 12.01 g/mol
The molar mass of O = 16.00 g/mol
So, the molar mass of CaCO3 = (40.08 + 12.01 + 3 * 16.00) g/mol
Now, we can calculate the number of moles of CaCO3 using the formula:
Number of moles = Mass / Molar mass
Number of moles of CaCO3 = 21.7 g / (40.08 + 12.01 + 3 * 16.00) g/mol
Step 2: Use the balanced chemical equation to find moles of CaCO3 that react
From the balanced chemical equation, we know that 1 mole of CaCO3 reacts to form 1 mole of CaO and 1 mole of CO2.
Therefore, the number of moles of CaCO3 that will react to reach equilibrium is equal to the number of moles of CaO formed, which is equal to the number of moles of CO2 formed.
Step 3: Calculate the percentage by mass
To determine the percentage by mass of CaCO3 that reacted, we'll compare the moles of CaCO3 that reacted to the initial moles of CaCO3.
Percentage by mass = (Moles of CaCO3 reacted / Initial moles of CaCO3) * 100
Finally, substitute the values we calculated into the equation to find the percentage by mass of CaCO3 that reacted.