Calculate the percent yield for a reaction where 0.38 g of NO2 reacts and 0.36 g N2O5 is produced.?
What mass of N2O5 will result from the reaction of 6.0 mol of NO2 if there is a 61.1% yield in the reaction?
I don't know how to work this out.please help. thank you!
Answer:
Okay, to calculate the percent yield you first must balance the equation.
2NO2 + O3 --> N2O5 + O2
you add a 2 coefficient to NO2 to balance the number of nitrogens and oxygens.
You next must find the theoretical yield-the most grams of N2O5 that could possibly be produced by this reaction. You convert the grams of NO2 to moles of NO2 and moles of N2O5 to grams of N2O5. NO2 is the limiting reactant as it requires 2 moles of NO2 to make one mole of N2O5.
.38g NO2 * 1 mol NO2/(14g +2*16 g) * 1 mol N2O5/2 mol NO2 * 108 g N2O5/ 1 mol N2O5 = 0.4461 g N2O5
This number is the maximum number of grams of N2O5 that can be produced by the reaction.
Divide the actual yield by the theoretical yield and multiply by 100 to get the percent yield.
(0.36 g N2O5/ 0.4461 g N2O5) * 100% = 80.7 % yield and the answer will be written as 81% yield to keep the same number of significant figures.
The next question is easier. Find the maximum theoretical yield again.
6 mol NO2 * 1 mol N2O5/2 mol NO2 * 108 g N2O5/1 mol N2O5 = 324 g N2O5
Multiply this by the percent yield
324 g N2O5 * .611 = 197.964g N2O5 and the answer with the correct significant figures is 198 g N2O5
Molecular weight NO2 = 46 g/mol
0.38 g / 46 = 0.00826 moles NO2
The balanced equation is
2 NO2 + O3 >> N2O5 + O2
the ratio between NO2 and N2O5 is 2 : 1
we would get 0.00826 /2 = 0.00413 moles N2O5
Molecular mass N2O5 = 108 g/mol
0.00413 mol x 108 g/mol = 0.446 g ( 100% yield)
0.36 : 0.446 = x : 100
x = 80.7%
6.0 moles NO2 would give 3.0 moles N2O5
3.0 x 108 = 324 g ( 100% yield)
61.1 : 100 = x : 324
x = 198 g mass N2O5
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