The compound is Al2O3. The ratio of aluminum to oxygen is 2:3.
Answer:
63.6 dm³
Explanation:
Volume of gas= no. of moles ×24dm³
Thus, volume of N₂O
= 2.65 × 24
= 63.6 dm³
Answer:
50
Explanation:
We will need a balanced equation with masses, moles, and molar masses of the compounds involved.
1. Gather all the information in one place with molar masses above the formulas and masses below them.
Mᵣ: 30.01 32.00 46.01
2NO + O₂ ⟶ 2NO₂
Mass/g: 80.00 16.00
2. Calculate the moles of each reactant

3. Calculate the moles of NO₂ we can obtain from each reactant
From NO:
The molar ratio is 2 mol NO₂:2 mol NO

From O₂:
The molar ratio is 2 mol NO₂:1 mol O₂

4. Identify the limiting and excess reactants
The limiting reactant is O₂ because it gives the smaller amount of NO₂.
The excess reactant is NO.
5. Mass of excess reactant
(a) Moles of NO reacted
The molar ratio is 2 mol NO:1 mol O₂

(b) Mass of NO reacted

(c) Mass of NO remaining
Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO
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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.