Answer:
41.11 g of Ca(OH)2.
Explanation:
The balanced equation for the reaction is given below:
CaO(s) + H2O(l) → Ca(OH)2(aq)
Next, we shall determine the masses of CaO and H2O that reacted and the mass of Ca(OH)2 produced from the balanced equation. This can be obtained as follow:
Molar mass of CaO = 40 + 16 = 56 g/mol
Mass of CaO from the balanced equation = 1 × 56 = 56 g
Molar mass of H2O = (2×1) + 16 = 2 + 16 = 18 g/mol
Mass of H2O from the balanced equation = 1 × 18 = 18 g
Molar mass of Ca(OH)2 = 40 + 2(16 + 1)
= 40 + 2(17)
= 40 + 34
= 74 g/mol
Mass of Ca(OH)2 from the balanced equation = 1 × 74 = 74 g
From the balanced equation above,
56 g of CaO reacted with 18 g of H2O to produce 74 g of Ca(OH)2.
Next, we shall determine the limiting reactant. This can be obtained as follow:
From the balanced equation above,
56 g of CaO reacted with 18 g of H2O.
Therefore, 33 g of CaO will react with = (33 × 18)/56 = 10.6 g of H2O.
From the calculation made above, we can see that it will take a higher amount (i.e 10.6 g) of H2O than was given (i.e 10 g) to react completely with 33 g of CaO. Therefore, H2O is the limiting reactant.
Finally, we shall determine the mass of Ca(OH)2 produced from the reaction.
In this case, the limiting reactant will be use because it will give the maximum yield of Ca(OH)2.
The limiting is H2O and the mass of Ca(OH)2 produced can be obtained as follow:
From the balanced equation above,
18 g of H2O reacted to produce 74 g of Ca(OH)2.
Therefore, 10 g of H2O will react to produce = (10 × 74)/18 = 41.11 g of Ca(OH)2.
Thus, 41.11 g of Ca(OH)2 were obtained from the reaction.
