2n2o5(g)→4no2(g)+o2(g) when 40g of n2o5 decomposes, 4.5g of o2 is formed. what is the percent yield

Answer 1

To solve this problem, we need to use stoichiometry and the concept of percent yield.

From the balanced chemical equation, we know that 1 mole of N2O5 produces 1 mole of O2 and 4 moles of NO2. Therefore, the theoretical yield of O2 can be calculated as follows:

n(N2O5) = m/M = 40/108 = 0.3704 mol (where m is the mass of N2O5 and M is its molar mass)

n(O2) = n(N2O5) x (1 mole of O2/1 mole of N2O5) = 0.3704 mol

m(O2) = n(O2) x M = 0.3704 x 32 = 11.85 g (where M is the molar mass of O2)

So, the theoretical yield of O2 is 11.85 g.

The actual yield of O2 is given as 4.5 g.

The percent yield can be calculated as:

% yield = (actual yield / theoretical yield) x 100%

% yield = (4.5 / 11.85) x 100% = 38.04%

Therefore, the percent yield of O2 is 38.04%.

Answer 2

This is in error. 2 moles N2O4 will produce 1 mole O2. The bots error is that 1 mol N2O4 will produce 1/2 mol O2; therefore, the percent yield calculated by the bot will be 2 x 38.04 assuming all of the other calculations are correct.

Answer 3

I apologize for the mistake in my previous response. You are correct that 2 moles of N2O5 will produce 1 mole of O2, and so the theoretical yield of O2 is:

n(O2) = n(N2O5) x (1 mole of O2/2 moles of N2O5) = 0.1852 moles

m(O2) = n(O2) x M = 0.1852 x 32 = 5.926 g

So, the theoretical yield of O2 is 5.926 g.

The actual yield of O2 is 4.5 g.

The percent yield can be calculated as:

% yield = (actual yield / theoretical yield) x 100%

% yield = (4.5 g / 5.926 g) x 100% = 76.0%

Therefore, the percent yield of O2 is 76.0%. Apologies for any confusion caused by my previous response.