Question

A sample of KCLO3 on decomposition yielded 448 ml of oxygen gas at NTP. Calculate weight of oxygen product .

Answer 1

Answer : The weight of oxygen is, 0.5856 gram.

Solution :

Using ideal gas equation,

$PV=nRT$

where,

At NTP,

P = pressure of the gas  = 1 atm

V = volume of the gas  = 448 ml = 0.448 L    (1 L = 1000 ml)

T = temperature of the gas  = 298.15 K

n = number of moles of gas

R = Gas constant = 0.0821 Latm/moleK

Now put all the given values in the above formula, we get

$n=\frac{PV}{RT}$

$n=\frac{(1atm)\times (0.448L)}{(0.0821Latm/moleK)\times (298.15K)}=0.0183moles$

Now we have to calculate the mass of oxygen.

The molar mass of Oxygen = 32 g/mole

$\text{Mass of oxygen}=\text{Moles of oxygen}\times \text{Molar mass of oxygen}$

$\text{Mass of oxygen}=0.0183moles\times 32g/mole=0.5856g$

Therefore, the weight of oxygen is, 0.5856 gram.

Answer 2

Answer:

The answer is :0.64g

Explanation:

One mol of an ideal gas occupies 22.4L volume.

1 mol O2-32g

THUS,1 mol O2(or 32g O2) occupy 22.4L volume.

Also,22.4L volume occupied by 32g O2

Thus,0.448L volume occupied by

(32/22.4 multiplied by 0.448)=0.64g.

Thanks