Asked by Adulkarim
Calculate the volume of oxygen starndad temperature and pressure which theoretically could be obtained from the 50cm3 of the solution of hydrogen peroxide containing 68g/dm3
Answers
Answered by
DrBob222
68 g x *50 cc/1000 dm^3 = ? g H2O2.
mols H2O2 = grams/molar mass.
Then you know 1 mol occupies 22.4L a STP.
mols H2O2 = grams/molar mass.
Then you know 1 mol occupies 22.4L a STP.
Answered by
ABEL
From; Concentration g/dm3=mass g/Volume(dm3)
So you can make the subject mass(g) =conc×volume
Mass of H2O2=68g/dm3×0.05dm3
Mass of H2O2= 3.4g
Then moles of H2O2= Mass/Molarmass
Moles of H2O2= 3.4g/34g/mol
Moles of H2O2=0.1mol
From the balanced decomposition reaction of hydrogen peroxide
2mol of H2O2 =1mol of oxygen
0.1mol of H2O2=? Mol of oxygen
= 0.05 moles of oxygen
At s.t.p number of moles of oxygen= v/V
Volume required=number of moles of oxygen ×V (s.t.p)
V=0.05mol ×22.4dm3
Volume of oxygen=1.12dm3
Ok abel yona john sagali
So you can make the subject mass(g) =conc×volume
Mass of H2O2=68g/dm3×0.05dm3
Mass of H2O2= 3.4g
Then moles of H2O2= Mass/Molarmass
Moles of H2O2= 3.4g/34g/mol
Moles of H2O2=0.1mol
From the balanced decomposition reaction of hydrogen peroxide
2mol of H2O2 =1mol of oxygen
0.1mol of H2O2=? Mol of oxygen
= 0.05 moles of oxygen
At s.t.p number of moles of oxygen= v/V
Volume required=number of moles of oxygen ×V (s.t.p)
V=0.05mol ×22.4dm3
Volume of oxygen=1.12dm3
Ok abel yona john sagali
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