pls help me out with this problem

Asked by  | 16th Jul, 2009, 07:04: PM

Expert Answer:

The reaction is :

2O3 3 O2

2 vol.               3  vol.

Let the volume of ozone in the initial mixrure be x mL

The volume of oxygen in the initial mixture will be (150-x) mL

Now, from the reaction we know that

2 volumes of ozone on decomposition will give 3 volume of oxygen.

x mL of ozone on decomposition will give (3/2)x  volume of oxygen.

Total volume of mixture after decomposition will be = (3/2)x  + (150 - x) mL

(3/2)x  + (150 - x) = 160

x = 20 mL

So,the volume of ozone in the initial mixrure be 20 mL

The volume of oxygen in the initial mixture will be (150-x) mL = 130 mL

To calculate the weight percentage of oxone in the original mixture:

22.4 L of O3 at STP will weigh 48 g

20 L of O3 at STP will weigh (48/ 22.4) x 20  g = 42.86 g

2.4 L of O2 at STP will weigh 32 g

130 L of O2 at STP will weigh (32/ 22.4) x 130  g = 185.71 g

Total mass of the mixture = 42.86 + 185.71 = 228.57 g

Weight percentage of ozone in the original mixture  = (42.86/ 228.57) x 100 = 18.75%

Answered by  | 10th Aug, 2009, 10:48: AM

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