How many grams of carbon dioxide will form if 4.4 g of propane reacts with 16.0 g of oxygen producing 7.2 g of water?
This is a limiting reagent. I know that because amounts are given for BOTH reactants.
C3H8 + 5O2 ==> 3CO2 + 4H2O
mols C3H8 = 4.4/molar mass = about 0.1.
mols O2 = 16/molar mass = about 0.5
Convert each to mols H2O, then to grams H2O.
0.1 mol C3H8 x (4 mols H2O/1 mol C3H8) = 0.4 mol H2O.
g = mols x molar mass = about 0.4 x 18 = 7.2 g which is the amount of water produced. This means we guessed right and propane is the limiting reagent.
Now convert mols C3H8 to mols CO2 then to g CO2.
0.1 mol C3H8 x (3 mols CO2/1 mol C3H8) = 0.1 x 3 = 0.3 mol CO2.
g CO2 = mols x molar mass.
Check all of those estimated numbers.
To determine the amount of carbon dioxide formed in the reaction, we need to first write the balanced chemical equation for the combustion of propane.
The balanced equation is:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
From the balanced equation, we can see that for every 1 mole of propane (C₃H₈) that reacts, we get 3 moles of carbon dioxide (CO₂) produced.
To find the number of moles of propane (C₃H₈), oxygen (O₂), and water (H₂O) involved in the reaction, we will use the formula:
moles = mass / molar mass
The molar mass of propane (C₃H₈) is:
3(C) + 8(H) = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol
moles of propane = 4.4 g / 44.11 g/mol = 0.1 mol
The molar mass of oxygen (O₂) is:
2(O) = 2(16.00 g/mol) = 32.00 g/mol
moles of oxygen = 16.0 g / 32.00 g/mol = 0.5 mol
Now, we need to find the limiting reactant, which is the reactant that is completely consumed and determines the amount of product formed. To do this, we compare the moles of propane and oxygen and see which one is present in the lowest amount.
From the balanced equation, we can see that 1 mole of propane (C₃H₈) reacts with 5 moles of oxygen (O₂). Therefore, to determine the moles of propane needed to react with the given amount of oxygen, we multiply the moles of oxygen by the ratio:
moles of propane needed = 5 * moles of oxygen = 5 * 0.5 mol = 2.5 mol
Since we only have 0.1 mol of propane, it is the limiting reactant. This means that all of the propane will react, and we will produce as much carbon dioxide as the balanced equation shows.
From the balanced equation, we know that 1 mole of propane (C₃H₈) produces 3 moles of carbon dioxide (CO₂).
moles of carbon dioxide = 3 * moles of propane = 3 * 0.1 mol = 0.3 mol
Finally, we can find the mass of carbon dioxide produced using the formula:
mass = moles * molar mass
The molar mass of carbon dioxide (CO₂) is:
1(C) + 2(O) = 1(12.01 g/mol) + 2(16.00 g/mol) = 44.01 g/mol
mass of carbon dioxide = 0.3 mol * 44.01 g/mol = 13.2 g
Therefore, 13.2 grams of carbon dioxide will form when 4.4 grams of propane reacts with 16.0 grams of oxygen, producing 7.2 grams of water.