Taking into account the reaction stoichiometry, the theoretical yield of oxygen when 29.2 g of water are decomposed by electrolysis is 25.956 grams.
<h3>Reaction stoichiometry</h3>In first place, the balanced reaction is:
2 H₂O → 2 H₂ + O₂
By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
The molar mass of the compounds is:
Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:
The theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.
<h3>Theorical yield of O₂</h3>The following rule of three can be applied: if by reaction stoichiometry 36 grams of water form 32 grams of oxygen, 29.2 grams of water form how much mass of oxygen?
<u><em>mass of oxygen= 25.956 grams</em></u>
Finally, the theoretical yield of oxygen when 29.2 g of water are decomposed by electrolysis is 25.956 grams.
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Answer:
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as
The equation for the ionization of the is given as:
When the solubility of is S moles/liter, then the solubility of
will be S moles/liter and solubility of
will be 2S moles/liter.
By stoichiometry of the reaction:
1 mole of gives 1 mole of
and 2 moles of
Thus for
is
Answer:
ΔH° (Enthalpy change) for the reaction is equal to (-151.5 KJ)
Explanation:
Determine the enthalpy for the reaction
COCl₂ (g) + H₂O(l) --> CH₂Cl₂(l) + O₂(g)
It becomes easier to find the unknown value of enthalpy of a particular reaction by applying the Hess's law on the given reaction with the known value of standard enthalpy change.
H₂(g) +
Cl₂(g) --> HCl(g) ΔH=-46 kJ -----------------------------(1)
H₂O(g) + Cl₂(g) --> 2 HCl(g) + O₂ (g) ΔH=-21 KJ-------------------(2)
CH₂Cl₂(l) + H₂(g) + O₂(g) --> COCl₂(g) + 2 H₂O(l) ΔH = 80.5 kJ ----------(3)
Equation (1) Multiply by factor 2 we get
H₂(g) + Cl₂(g) ---> 2 HCl (g) ΔH = - 46 x 2 = - 92 KJ ------------(4)
Revers the equation (2) and (3) we get
2 HCl(g) + O₂ (g) --> H₂O(g) + Cl₂(g) ΔH=+21 KJ ----------------(5)
COCl₂(g) + 2 H₂O(l) --> CH₂Cl₂(l) + H₂(g) + O₂(g) ΔH = - 80.5 kJ -----------(6)
Now add these three equations i.e., (4) + (5) + (6)
we can get these equation
COCl₂ (g) + H₂O(l) --> CH₂Cl₂(l) + O₂(g)
The enthalpy change during the reaction = -92 + 21 -80.5 = - 151.5 KJ