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.
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?
1 answer