N2(g) + O2(g) --> N2O5(g)
(a) Consider the unbalanced equation above. What is the maximum mass of N2O5 that can be produced when 43.0 g of N2 and 43.5 g O2 react?
Another limiting reagent problem Jade. All of these are one the same way.
To determine the maximum mass of N2O5 that can be produced, we need to follow these steps:
Step 1: Write the balanced chemical equation.
N2(g) + O2(g) --> N2O5(g)
Step 2: Calculate the molar masses (grams/mole) of N2, O2, and N2O5.
Molar mass of N2: 28.02 g/mol
Molar mass of O2: 32.00 g/mol
Molar mass of N2O5: 108.01 g/mol
Step 3: Determine the limiting reactant.
The limiting reactant is the one that will be completely consumed, thus determining the maximum amount of product that can be formed. To find the limiting reactant, we need to compare the number of moles of each reactant.
First, convert the given masses of N2 and O2 to moles using their respective molar masses:
Moles of N2 = 43.0 g / 28.02 g/mol ≈ 1.536 mol
Moles of O2 = 43.5 g / 32.00 g/mol ≈ 1.359 mol
The ratio between N2 and N2O5 in the balanced equation is 1:1. Therefore, the reaction uses an equal number of moles of N2 and N2O5.
Using the mole ratios, we identify the limiting reactant by comparing the moles of N2 and O2:
N2: 1.536 mol
O2: 1.359 mol
Since we have fewer moles of O2, it is the limiting reactant.
Step 4: Determine the maximum mass of N2O5 that can be produced.
Now we need to calculate the maximum mass of N2O5 that can be produced using the moles of O2 (the limiting reactant).
Moles of N2O5 = Moles of O2 (from step 3) = 1.359 mol
Mass of N2O5 = Moles of N2O5 × Molar mass of N2O5
Mass of N2O5 = 1.359 mol × 108.01 g/mol ≈ 146.82 g
Therefore, the maximum mass of N2O5 that can be produced is approximately 146.82 grams.