Respuesta :
Answer:
22.52 liters volume of pure oxygen gas will be collected at 27°C and 764 Torr.
Explanation:
Mass of hydrogen peroxide solution = 125 g
Mass of hydrogen peroxide in solution = 50% of 125 g ;
[tex]\frac{50}{100}\times 125 g=62.5 g[/tex]
Moles of hydrogen peroxide =[tex]\frac{62.5 g}{34 g/mol}=1.838 mol[/tex]
[tex]2H_2O_2(aq)\rightarrow 2H_2O(l)+O_2(g)[/tex]
According to reaction, 2 moles of hydrogen peroxide gives 1 mole of oxygen gas.
Then 1.838 moles of hydrogen peroxide will give:
[tex]\frac{1}{2}\times 1.838 mol=0.919 mol[/tex] of oxygen gas .
Moles of oxygen gas = n = 0.919 mol
Pressure of oxygen gas , P= 764 Torr= [tex]\frac{764}{760}atm=1.005 atm[/tex]
Volume of oxygen gas = V = ?
Gas constant , R= 0.0821 L.atm/mol.K
Temperature of oxygen gas = 27°C=27+273 K=300 K
Using ideal gas equation:
PV = nRT
Putting values in above equation, we get:
[tex]V=\frac{nRT}{P}[/tex]
[tex]=\frac{0.919 mol\times 0.0821 atm L/mol K\times 300 K}{1.005 atm}[/tex]
V = 22.52 L
22.52 liters volume of pure oxygen gas will be collected at 27°C and 764 Torr.
