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.
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
4 answers
help
A
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%.
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%.