Ammonia undergoes combustion with oxygen to produce nitric oxide and water. The volume of the oxygen required to react with 720 ml of ammonia is 900 ml.
<h3>What is volume?</h3>
Volume is the area occupied by the substance and is the ratio of the mass to the density.
At STP, 1 mole of gas occupies 22.4 L of volume
Given,
Volume of ammonia reacted = 0.720 L
The combustion reaction is shown as,

From the stoichiometry of the reaction, it can be said that,
L of ammonia reacts with
L of oxygen gas.
So, 0.720 L of ammonia will react with:

Therefore, the volume of oxygen required is 900 mL.
Learn more about volume here:
brainly.com/question/14090111
#SPJ4
Answer is: sulfuric acid is the limiting reactant.
Chemical reaction: 3H₂SO₄ + 2Al(OH)₃ → Al₂(SO₄)₃ + 6H₂O.
m(H₂SO₄) = 34 g.
n(H₂SO₄) = m(H₂SO₄) ÷ M(H₂SO₄).
n(H₂SO₄) = 34 g ÷ 98 g/mol.
n(H₂SO₄) = 0,346 mol.
m(Al(OH)₃) = 33 g.
n(Al(OH)₃) = 33 g ÷ 78 g/mol.
n(Al(OH)₃) = 0,423 mol.
From chemical reaction: n(H₂SO₄) : n(Al(OH)₃) = 3 : 2.
This reaction is decomposition. It is the breakdown of a compound into simpler and smaller elements.
Answer:
0.4 M
Explanation:
The process that takes place in an aqueous K₂HPO₄ solution is:
First we <u>calculate how many K₂HPO₄ moles are there in 200 mL of a 0.2 M solution</u>:
- 200 mL * 0.2 M = 40 mmol K₂HPO₄
Then we <u>convert K₂HPO₄ moles into K⁺ moles</u>, using the <em>stoichiometric coefficients</em> of the reaction above:
- 40 mmol K₂HPO₄ *
= 80 mmol K⁺
Finally we <em>divide the number of K⁺ moles by the volume</em>, to <u>calculate the molarity</u>:
- 80 mmol K⁺ / 200 mL = 0.4 M