Question

1) If 1.63 g of sodium peroxide (Na₂O₂) react with water to produce sodium hydroxide and oxygen, how many liters of oxygen will be produced at 25.0 °C and 725 torr?

2 Na₂O₂(s) + 2 H₂O(l) → 4 NaOH(aq) + O₂(g)

Liters of oxygen =

2) A gas occupies a volume of 5.87 L at 0.950 atm. At what pressure will the volume be 8.45 L?

Pressure =

Answer 1

Soletion * 1) - Balanced chernical equation 2N4₂0215) + 2Hoc1) 4 N904 caps + O2(g) Given, mass of NGO2 = 1.63g Temperature = 25°c =25+2731= 298 Presszise cp2 - 725 torr = - 760tour latm - 725 x = torr +25 = 0.954 eetm R = 0.08206 Latm/molok - Hese From the balanced eqriation, 2 mole of NaO2 produce 1 mole of Oz =) Since, mole = mass Moler mass So, moles of N420 = 1.639 = 0.02ogmol 77.98 N9₂02 =) Hence, from the balanced equation moles of Na, 0₂ = 0.0209 X2 = 0.0418 mo) Nez@z moles of Oz prodriced = 0.0209 molą =) Hence

- Now from ideal gas eqrration PV = nRT V = RRT P = 0.02ogmolxo.08206 Latm/markt 2988 0.954 atm = 0.54€ z gas produced Liters of oxygen = 0.54L/ = Riven, V=5087L P = 0.950 etm 1₂ = 8.45L P2= ? By using Boyle's low equation PV = P2 V₂ P2= PIXI = 0.950 atmix 5.874 2 8.454 = 0.659 atm Henee, Pressuse = 0.659 atom