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using the actual moles of o2 you determined from your experiment (n) and the theoretical moles of o2 you just calculated, show your calculations from experiment 1: ideal gas law - finding percent h2o2 with yeast for determining the percent hydrogen peroxide in your experimental sample.

Answer :

Theoretical mass (O₂)= H₂O₂ Volume×  H₂O₂ density× {( 1 mol O₂)/(2 mol H2O2)}×{( mol of  H₂O₂ /g  H₂O₂ )}

H2O2 Volume= 5 ml

Known density( H₂O₂ )= 1.02 g/ml

Molar mass ( H₂O₂ )= 34.01 g/mol

Reciprocal molar mass(  H₂O₂ )= 0.0294

Theoretical mass( O₂)= 5×1.02× 0.0294× 1/2= 0.0749 moles

Actual moles of O₂ and percentage of  H₂O₂

P= 753 mm Hg×1 atm/760 mm Hg= 0.991 atm

V= 45  ml× 1L /1000 ml= 0.045 L

n= PV/RT

Actual moles ( O₂)= (0.991×0.045)/(0.0821×296 K)= 0.0445/24.3016= 1.8×10^-3 moles

Percentage ( H₂O₂ )= Theoretical moles(O₂)×100= 0.0749×100= 7.49%

Mass of  H₂O₂ (percent)= 3%

Concentration by mass( H₂O₂ )= 3 gm H2O in 100 ml of H2O

Volume ( H₂O₂ )= 100 ml

1 mol ( H₂O₂ )= 34.02 g O2

(3g  H₂O₂ /100 ml Solution)×( 1 mol  H₂O₂ /34.02 g H2O2)

No. Of moles ( H₂O₂ )= 0.088 moles

Mass ( H₂O₂ )= 0.088 moles × 34.02 = 2.99 g

%  H₂O₂ = (2.99 g  H₂O₂ /100 ml Solution)×100= 2.99%

% error= (%  H₂O₂ from the bottle-experiment %  H₂O₂ )/(% H₂O₂ from bottle)×100

=[ (3-2.99)/(3) ]× 100= 0.34% error.

Learn more about H₂O₂ here:- https://brainly.com/question/25566753

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