CaO ( calcium oxide) is a basic oxide.
Considering the reaction stoichiometry, the quantity of moles of H₂O produced is 3.02 moles.
The balanced reaction is:
CuO + H₂ → Cu + H₂O
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:
- CuO= 1 mole
- H₂= 1 mole
- Cu= 1 mole
- H₂O= 1 mole
The molar mass of the compounds present in the reaction is:
- CuO= 79.55 g/mole
- H₂= 2 g/mole
- Cu= 63.55 g/mole
- H₂O: 18 g/mole
Then, by reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of mass of each compound participate in the reaction:
- CuO= 1 mole× 79.55 g/mole= 79.55 grams
- H₂= 1 mole× 2 g/mole= 2 grams
- Cu= 1 mole× 63.55 g/mole= 63.55 grams
- H₂O: 1 mole× 18 g/mole= 18 grams
Then, it is possible to determine the moles of H₂O necessary by the following rule of three: If by reaction stoichiometry 79.55 grams of CuO produces 1 mole of H₂O, 240 grams of CuO produces how many moles of H₂O?

<u><em>moles of H₂O= 3.02 moles</em></u>
Finally, the quantity of moles of H₂O produced is 3.02 moles.
Learn more:
The percentage yield is 72.8 %.
<em>Step 1</em>. Calculate the <em>mass of Br₂</em>
Mass of Br₂ = 20.0 mL Br₂ × (3.10 g Br₂/1 mL Br₂) = 62.00 g Br₂
<em>Step 2</em>. Calculate the <em>theoretical yield</em>
M_r: 159.81 266.69
2Al + 3Br₂ → 2AlBr₃
Moles of Br₂ = 62.00 g Br₂ × (1 mol Br₂/(159.81 g Br₂) = 0.3880 mol Br₂
Moles of AlBr₃ = 0.3880 mol Br₂ × (2 mol AlBr₃/(3 mol Br₂) = 0.2586 mol AlBr₃
Theor. yield of AlBr₃ = 0.2586 mol AlBr₃ × 266.99 g AlBr₃)/(1 mol AlBr₃)
= 69.05 g AlCl₃
<em>Step 3</em>. Calculate the <em>percentage yield
</em>
% yield = (actual yield/theoretical yield) × 100 % = (50.3 g/69.05 g) × 100 %
= 72.8 %