Solid sodium peroxide (Na2O2) reacts with carbon dioxide gas to form solid sodium carbonate and oxygen gas. If I have 56.2 g of Na2O2 and I treat it with 2.4 L of carbon dioxide at 675 Torr and 23.6 „aC, how many grams of sodium carbonate will be produced?
This is a limiting reagent (LR) problem. You know that because an amount is given for BOTH reactants. I work these the long way.
Na2O2 + CO2 ==> Na2CO3 + O2
Use PV = nRT and solve for n = number of mols CO2 at the conditions listed.
mols Na2O2 = grams/molar mass
Use the coefficients in the balanced equation to convert mols CO2 to mols Na2CO3.
Do the same and convert mols Na2O2 to mols Na2CO3.
It is likely that the two values will not agree; the correct value in LR problem is ALWAYS the smaller value.
Using the smaller value, convert mols Na2CO3 to g. g = mols x molar mass = ?
To determine the grams of sodium carbonate produced, we first need to balance the chemical equation for the reaction:
2 Na2O2 + 2 CO2 -> 2 Na2CO3 + O2
From the balanced equation, we can determine the stoichiometric ratio between Na2O2 and Na2CO3. For every 2 moles of Na2O2, we get 2 moles of Na2CO3.
Next, we calculate the moles of Na2O2:
Molar mass of Na2O2 = (2 x 22.99 g/mol) + (2 x 16.00 g/mol) = 77.98 g/mol
Moles of Na2O2 = mass / molar mass = 56.2 g / 77.98 g/mol
Now we convert the moles of Na2O2 to moles of Na2CO3 using the stoichiometric ratio:
Moles of Na2CO3 = Moles of Na2O2 x (2 moles Na2CO3 / 2 moles Na2O2)
Next, we can calculate the volume of carbon dioxide gas in moles using the ideal gas law:
PV = nRT
n = PV / RT
n = (675 Torr * 2.4 L) / (0.0821 L*atm/(mol*K) * (273.15°C + 23.6°C)
Finally, we convert moles of CO2 to moles of Na2CO3 using the stoichiometric ratio:
Moles of Na2CO3 = Moles of CO2 x (2 moles Na2CO3 / 2 moles CO2)
Now that we have the moles of Na2CO3, we can calculate the grams of Na2CO3:
Grams of Na2CO3 = moles of Na2CO3 x molar mass of Na2CO3
By following these steps, you can determine the grams of sodium carbonate produced in the reaction.