Chemical Engineering - Stoichiometry - Discussion

The decomposition of water takes place in the following way -

H2O ⟶H2 + O2.

We find that the ratio of the mass of hydrogen to that of oxygen in water is 1:8, so when water decomposes it leads to the production of hydrogen and oxygen in 1:8 proportion.

So, this means that the decomposition of 90 kg of water produces 10 kg of hydrogen and 80 kg of oxygen and the total mass before and after decomposition is 90 kg.

This follows the law of constant proportions.

Thus 80 kg of oxygen can be produced by 90 kg of water.

Please give me the procedure.

Please explain the answer.

Please explain the answer.

Please explain the answer clearly.

H2+1/2O2=H2O.
2+32/2=18.

18H2O=08O2,
So,
90H2O=16/18*90. O2.
O2= 16*5 =80 kg O2.

I am not getting it, please explain the answer.

(16/18) * 90 = 80.

H2+1/2O2---> H2O.

2+1/2 * 32=18.

Now taking
H2O:O2,
18:16,
1:16/18,
90:16/18 * 90,
= 80.

H2+2O2 = 2H20.

90 grams of water = 5 moles of water (moles=weight/molecular weight)

That means 2 moles of O2 = 2 moles of H2o,
?moles of O2 = 5 moles of H2O,
5x2÷2 = 5 moles.
Convert moles of O2 to the weight of O2 get 80 gms.