Oxygen (6O2) and Glucose (C6H12O6)
<span>Reference: 6CO2 + 6H2O + light energy = C6H12O6 + 6O2.</span>
The volume of H₂O = 5 L
<h3>Further explanation</h3>
Given
5L of H₂ and 3L O₂
Reaction
2H₂ (g) + O₂(g) ⇒2H₂O(g)
Required
The volume of H₂O
Solution
Avogadro's hypothesis:
<em>In the same T,P and V, the gas contains the same number of molecules </em>
So the ratio of gas volume will be equal to the ratio of gas moles
mol H₂ = 5, mol O₂ = 3
From equation, mol ratio H₂ : O₂ = 2 : 1, so :
mol H₂O based on mol H₂, and from equation mol ratio H₂ : H₂O=2 : 2, so mol H₂O = 5 mol and the volume also 5 L
We have as a reagent a salt, lead nitrate (Pb(NO3)2), and an unknown solution that gives us as a product lead chloride (PbCl2). That is, the solution must contain chlorine.
If a chlorine solution is used we will have the following reaction:
So, with a chlorine solution, we will have a white precipitate of lead chloride.
Answer: 2. I see two colours in the test tube, white and grey at the bottom.
Explanation:
Balance between K+ charge and Br- charge = 0