Since
21.2 g H2O was produced, the amount of oxygen that reacted can be obtained
using stoichiometry. The balanced equation was given: 2H₂ + O₂ → 2H₂O and
the molar masses of the relevant species are also listed below. Thus, the
following equation is used to determine the amount of oxygen consumed.
Molar mass of H2O = 18
g/mol
Molar mass of O2 = 32
g/mol
21.2 g H20 x 1 mol
H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2
<span>We then determine that
18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is
important to note that we do not need to consider the amount of H2 since we can
derive the amount of O2 from the product. Additionally, the amount of H2 is in
excess in the reaction.</span>
Limitations of Van der waal's equation. (i) The value of 'b' is not constant but varies with pressure and temperature. (ii) The value of is not equal to 3b, but actually it is equal to, in some case; and in other cases 2b. (iii) The value of is not equal to but it is usually more than 3 for most of the gases.
The number of moles in 32.5g of aluminum chloride is approximately 0.250 moles.
Answer:

Explanation:
10.11 g sample of
contains 22.34%
by mass
According to the law of constant composition the if one sample of
has 22.34% of
by mass then any other sample of
will have the same percentage of the amount of
.
For a sample of 7.09 g we have

The mass of sodium in the required sample is
.