Question

Calculate the volume and no. Of moles of oxygen liberated at stp when 5.2gm of sodium peroxide reacts with water.
2Na2O2+2H2O→4NaOH+O2

Answer 1

Explanation:

No of moles of Sodium peroxide= mass/Mr

= 5.2 / 78

= 0.067 mol (3sf)

Sodium peroxide : Oxygen

2 : 1

No of moles of Oxygen = 0.067/2

= [0.0335 moles]

Vol of Oxygen = no of moles x 24

= 0.0335 x 24

= [0.804 dm^3]

Answer 2

The volume and no. of moles of oxygen liberated at STP when 5.2gm of sodium peroxide reacts with water is 0.78 L and 0.035 moles

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP, weighs equal to the molecular mass and contains avogadro's number  $6.023\times 10^{23}$ of particles.

moles of $Na_2O_2=\frac{\text {given mass}}{\text {molar mass}}=\frac{5.2g}{78g/mol}=0.07mol$

$2Na_2O_2+2H_2O\rightarrow 4NaOH+O_2$

According to stoichiometry:

2 moles of $Na_2O_2$ gives 1 mole of $O_2$

0.07 moles of $Na_2O_2$ gives =$\frac{1}{2}\times 0.07=0.035$  moles of $O_2$

Standard condition of temperature (STP)  is 273 K and atmospheric pressure is 1 atmosphere respectively.

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = 1 atm (at STP)

V= Volume of the gas = ?

T= Temperature of the gas = 273 K  (at STP)  

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= 0.035

$V=\frac{nRT}{P}=\frac{0.035\times 0.0821\times 273}{1atm}=0.78L$

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