Answer:
41.11 g of Ca(OH)2.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CaO + H2O —> Ca(OH)2
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 = (2x1) + 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.
Finally, we obtained the mass of calcium hydroxide, Ca(OH)2 produced from the reaction of 33 g of CaO and 10 g of H2O. This can be obtained as follow:
From the question given above, we were told that when 33 g of CaO and 10 g of H2O reacted, 2 g of CaO were remaining. This implies that H2O is the limiting reactant and CaO is the excess reactant.
Thus, we shall use the limiting reactant to determine the mass of calcium hydroxide, Ca(OH)2 produced because it will give the maximum yield as all of it is consumed in the reaction.
The limiting reactant is H2O and the mass of calcium hydroxide, 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.
Therefore, 41.11 g of Ca(OH)2 were obtained from the reaction.
(atoms of C in one mole) = 6.84*