Answer: D (The abundance percentage of each isotope)
Explanation: hope this helps!
Taking into account the reaction stoichiometry, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.
<h3>Reaction stoichiometry</h3>
In first place, the balanced reaction is:
CaO + H₂O → Ca(OH)₂
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:
- CaO: 1 mole
- H₂O: 1 mole
- Ca(OH)₂: 1 mole
<h3>Moles of CaO required</h3>
The following rule of three can be applied: If by stoichiometric reaction 1 mole of Ca(OH)₂ is produced by 1 mole of CaO, 2 moles of Ca(OH)₂ are produced by how many moles of CaO?

moles of CaO= 2 moles
Finally, 2 moles of CaO are required to react with 2 moles of Ca(OH)₂.
Learn more about the reaction stoichiometry:
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Answer:
72.53% is the yield of CrCl3
Explanation:
Given
Reaction:
Cr2O3(s) + 3 CCl4(l) → 2 CrCl3(s) + 3 COCl2(aq)
CCl4 is in excess and 17.6g Cr2O3 present
The reaction yields 26.6g of CrCl3
To Find:
% yields of the reaction
Also given
Molar mass of CrCl3 = 158.35g/mol
Molar mass of Cr2O3 = 152.00 g/mol
By the stoichiometry of the reaction
1 mole of Cr2O3 gives 2 moles of CrCl3
0r
1 x1 52 g of Cr2O3 gives 2x 158.35 g of CrCl3
= 1 52 g of Cr2O3 gives 316.70 g of CrCl3
17.6 g of Cr2O3 gives (17.6÷152) × 316.70 g CrCl3
= 36.67 g CrCl3
but actual yield is only 26.6g
so % yield is (26.6 ÷÷ 36.67) × 100
= 72.53% is the yield of CrCl3