How many litres (L) of oxygen at STP can be obtained from 110 g of potassium chlorate? [R=0.0821 L.atm/K.mol]

Chemistry

1 answer:

lisabon 2012 [21]3 years ago

3 0

Balance the equation first:

2 KClO3 (s) ---> 2 KCl (s) + 3 O2 (g)

Moles of KClO3 = 110 / 122.5 = 0.89

Following the balanced chemical equation:
We can say moles of O2 produce = $\frac{3}{2}$ x moles of KClO3

So, O2 = (3 / 2) x 0.89

= 1.34 moles

So, Volume at STP = nRT / P

T = <span>273.15 K
P = 1 atm

So, V = (1.34 x 0.0821 x 273.15) / 1 = 30.2 L</span>

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<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.

Usually radioactive elements have an unstable atomic nucleus.

General formulas used in decay:

$\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}$