Question

What is the percentage yield of N2O5 when 68.0g of O2 reacts With N2 according to the reaction equation, with 72.6g of N2O5 being obtained?
2N2 + 5O2 -> 2N2O5

Answer 1

I assume that is an excess of N2.
2N2 + 5O2 -> 2N2O5
Mols O2 = grams/molar mass = 68.0/32 = approx 2
mols N2O5 formed if 100% yield = approx 2 x (2/5) = about 0,8.
grams N2O5 = mols N2O5 x molar mass N2O4 = ? This is the theoretical yield(TY).
The actual yield (AY) is 72.6 g
% yield = (AY/TY)*100 = ?
Check my work. Remember all of the numbers I've shown are estimates and must be recalculated from scratch.

Answer 2

help

Answer 3

A

Answer 4

Alright, let's break down this problem step by step.

1. Determine the moles of O2:

Moles O2 = mass ÷ molar mass = 68.0 g ÷ 32 g/mol = 2.125 mol

2. Use the mole ratio between O2 and N2O5 to determine the theoretical yield of N2O5:

From the balanced chemical equation, we can see that for every 5 moles of O2, we should get 2 moles of N2O5. Therefore,

Moles N2O5 (from O2) = moles O2 × (2 mol N2O5 ÷ 5 mol O2)

Moles N2O5 = 2.125 mol × (2/5) = 0.85 mol

3. Calculate the theoretical yield of N2O5:

The molar mass of N2O5 is 108 g/mol, so

Theoretical yield = moles N2O5 × molar mass N2O5 = 0.85 mol × 108 g/mol = 91.8 g

4. Calculate the percent yield:

Percent yield = actual yield ÷ theoretical yield × 100%

Percent yield = 72.6 g ÷ 91.8 g × 100% = 79.1%

Therefore, the percentage yield of N2O5 is approximately 79.1%.