How much oxygen gas is produced in L if the reaction occurs at 34.4 ºC and 1.00 atm pressure?

Question

A person accidentally mixes bleach (sodium hypochlorite solution) with hydrogen peroxide solution. The following reaction occurs and produces oxygen gas. 18.8 grams of NaClO were present in the bleach, while 21.4 grams of H₂O₂ were present in the hydrogen peroxide solution. How much oxygen gas is produced in L if the reaction occurs at 34.4 ºC and 1.00 atm pressure?

H₂O₂ (aq) + NaClO (aq) → NaCl (aq) + H₂O (aq) + O₂ (g)

Answer 1

H₂O₂ (aq) + NaClO (aq) ==> NaCl (aq) + H₂O (aq) + O₂ (g) ... balanced equation

moles H2O2 present = 21.4 g x 1 mol H2O2/34 g = 0.629 moles

moles NaClO present = 18.8 g x 1 mol NaClO/74.5 g = 0.252 moles

Since they react in a 1:1 mol ratio, NaClO is LIMITING and it determines moles of O2 produced.

Moles O2 produced = 0.252 mol NaClO x 1 mol O2/mol NaClO = 0.252 moles O2 prouced

Volume O2 produced:

PV = nRT

V = nRT/P = (0.252 mol)(0.0821 Latm/Kmol)(307K)/1 atm

V = 6.36 L

Answer 2

34g of H2O2 will react with 74.5g of NaOCl to produce 32g of oxygen.

74.5g of NaOCl would give 32g of Oxygen

So, 18.8g of NaOCl would react with 8.5g of H2O2 to produce 32/74.5X18.8g = 8.075g of oxygen

PV=nRT

1.V=w/M0.0821X307

V= (8.075/32)X0.0821X307

=6.36L