4.4 g of CaO reacts with 7.77 g of water, as shown in the following balanced chemical equation:
CaO(s) + H2O (l) →Ca(OH)2(aq)
a)Identify the limiting reagent.
b)How many grams of calcium hydroxide will be formed?
CaO(s) + H2O (l) →Ca(OH)2(aq)
There are two ways to determine the limiting reagent(LR). Here is the easy way but I'll be glad to show you the long way if you wish.
Step 1: Determine the moles of each reagent.
mols CaO = g/molar mass = 4.4 g/56 = 0.079
mols H2O = 7.77/18 = 0.43
Step 2: Take one at a time.
If CaO is the LR, you will need 0.079 mols of H2O You have that much; therefore CaO is the LR. But check it with H2O.
If H2O is the LR you will need 0.43 mols CaO and you don't have that much so H2O can't be the LR.
2. You have 0.079 mols CaO. Since you get 1 mol Ca(OH)2 for every 1 mol CaO initially, then you will form 0.079 mols Ca(OH)2 formed. Grams = mols x molar mass = ? = theoretical yield @100% effeciency.
Post your work if you get stuck.
To determine the limiting reagent, we need to compare the moles of CaO to the moles of H2O.
a) Moles of CaO = mass of CaO / molar mass of CaO
= 4.4 g / 56.08 g/mol (molar mass of CaO)
= 0.0784 mol
Moles of H2O = mass of H2O / molar mass of H2O
= 7.77 g / 18.015 g/mol (molar mass of H2O)
= 0.431 mol
Based on the equation, the stoichiometry ratio between CaO and H2O is 1:1.
Therefore, the limiting reagent is the one that has the smallest number of moles, which in this case is CaO.
b) To find the amount of calcium hydroxide formed, we need to use the stoichiometry of the balanced equation. From the equation, we can see that for every 1 mole of CaO, 1 mole of Ca(OH)2 is formed.
Moles of Ca(OH)2 formed = Moles of CaO = 0.0784 mol
Mass of Ca(OH)2 formed = Moles of Ca(OH)2 formed * Molar mass of Ca(OH)2
= 0.0784 mol * 74.09 g/mol (molar mass of Ca(OH)2)
= 5.81 g
Therefore, 5.81 grams of calcium hydroxide will be formed during the reaction.
To identify the limiting reagent, we need to compare the amounts of reactants given and determine which one will run out first, limiting the amount of product that can be formed.
a)
Step 1: Convert the given masses of reactants into moles by using their molar masses.
Molar mass of CaO:
Ca: 40.08 g/mol
O: 16.00 g/mol
Total molar mass of CaO: 40.08 + 16.00 = 56.08 g/mol
Moles of CaO = mass of CaO / molar mass of CaO
Moles of CaO = 4.4 g / 56.08 g/mol ≈ 0.078 mol
Molar mass of H2O:
H: 1.01 g/mol
O: 16.00 g/mol
Total molar mass of H2O: 1.01 + 16.00 = 17.01 g/mol
Moles of H2O = mass of H2O / molar mass of H2O
Moles of H2O = 7.77 g / 17.01 g/mol ≈ 0.456 mol
Step 2: Calculate the mole ratio of CaO to H2O using the balanced chemical equation.
From the balanced equation: 1 mol CaO reacts with 1 mol H2O
The mole ratio of CaO to H2O is 1:1.
Step 3: Compare the moles of CaO and H2O to identify the limiting reagent.
Since the mole ratio is 1:1, the limiting reagent is the one with fewer moles. In this case, CaO has 0.078 mol and H2O has 0.456 mol. Therefore, CaO is the limiting reagent.
b)
Step 1: Calculate the moles of Ca(OH)2 formed using the mole ratio of CaO to Ca(OH)2 from the balanced chemical equation.
Mole ratio from the balanced equation: 1 mol CaO produces 1 mol Ca(OH)2
Moles of Ca(OH)2 = Moles of CaO = 0.078 mol (since CaO is the limiting reagent)
Step 2: Calculate the mass of Ca(OH)2 formed by multiplying the moles of Ca(OH)2 by its molar mass.
Molar mass of Ca(OH)2:
Ca: 40.08 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Total molar mass of Ca(OH)2: 40.08 + 2(1.01) + 16.00 = 74.17 g/mol
Mass of Ca(OH)2 = Moles of Ca(OH)2 * Molar mass of Ca(OH)2
Mass of Ca(OH)2 = 0.078 mol * 74.17 g/mol ≈ 5.78 g
Therefore, approximately 5.78 grams of calcium hydroxide (Ca(OH)2) will be formed.