Consider the following chemical reaction 2HCL+ Ca(OH)2 CaCl2+2H2O
You have mixed 30.00ml of 0.150 M HCl solution with 20.00ml of 0.100m Ca(OH)2 solution what is the molarity of the HCl, Ca( OH)2 and CaCl2 after the reaction has stopped
This is a limiting reagent (LR) problem with a twist or two.
mols HCl = M x L = about 0.0045
mols Ca(OH)2 = M x L = about 0.002
mols CaCl2 formed if we start with 0.0045 mols HCl and all of the Ca(OH)2 we need is
0.0045 mols HCl x (1 mol CaCl2/2 mols HCl) = 0.0045 x (1/2) = about 0.00225.
mols CaCl2 formed if we started with 0.002 mols Ca(OH)2 and all of the HCl we needed is
0.002 x (1 mol CaCl2/1 mol Ca(OH)2) = 0.002.
You have two different values for mols CaCl2 which means one of them is wrong. The correct value in LR problems is ALWAYS the smaller value and the reagent producing that value is the LR.
So all of the Ca(OH)2 will be used and you will form 0.002 mol CaCl2. Thus
[Ca(OH)2] = 0
[CaCl2] = 0.002 mols/0.05L = ?
For HCl we must determine how much of it has been used and how much is left.
mols used = 0.002 mol Ca(OH)2 x (2 mol HCl/1 mol Ca(OH)2) = 0.002 x 2 = 0.004.
We started with 0.0045 so we have left the difference or 0.0005 mols and that divided by the volume (in L) is the M.
By the way, you need to find and use the arrow key. You can make arrows either --> or ==>. And note to be careful with the M and m since they mean different things.
To find the molarity of HCl, Ca(OH)2, and CaCl2 after the reaction has stopped, we need to use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation for the reaction is:
2HCl + Ca(OH)2 -> CaCl2 + 2H2O
Given:
Volume of HCl solution = 30.00 mL
Concentration of HCl solution = 0.150 M
Volume of Ca(OH)2 solution = 20.00 mL
Concentration of Ca(OH)2 solution = 0.100 M
Step 1: Convert the given volumes to liters.
30.00 mL = 30.00 mL * (1 L / 1000 mL) = 0.03000 L
20.00 mL = 20.00 mL * (1 L / 1000 mL) = 0.02000 L
Step 2: Calculate the moles of HCl and Ca(OH)2.
Moles of HCl = Volume of HCl solution (L) * Concentration of HCl (M)
= 0.03000 L * 0.150 M
= 0.00450 moles
Moles of Ca(OH)2 = Volume of Ca(OH)2 solution (L) * Concentration of Ca(OH)2 (M)
= 0.02000 L * 0.100 M
= 0.00200 moles
Step 3: Determine the limiting reactant. Since the stoichiometry of the balanced equation is 2:1 for HCl and Ca(OH)2, the reactant that produces fewer moles will be the limiting reactant. In this case, Ca(OH)2 is the limiting reactant because it produces half as many moles as HCl.
Step 4: Use the stoichiometry of the balanced equation to determine the moles of CaCl2 formed.
From the balanced equation, we know that 2 moles of HCl react with 1 mole of Ca(OH)2 to produce 1 mole of CaCl2.
Therefore, moles of CaCl2 formed = (0.00200 moles of Ca(OH)2) * (1 mole of CaCl2 / 1 mole of Ca(OH)2)
= 0.00200 moles
Step 5: Calculate the new volumes of CaCl2 and H2O formed.
Since 2 moles of HCl react to produce 1 mole of CaCl2, the number of moles of HCl remaining after the reaction will be twice the moles of CaCl2 formed.
Moles of HCl remaining = 2 * moles of CaCl2
= 2 * 0.00200 moles
= 0.00400 moles
From the balanced equation, we know that 1 mole of CaCl2 reacts to produce 2 moles of H2O.
Therefore, moles of H2O formed = (0.00200 moles of CaCl2) * (2 moles of H2O / 1 mole of CaCl2)
= 0.00400 moles
Step 6: Calculate the new concentrations of HCl, Ca(OH)2, and CaCl2.
Molarity of HCl = Moles of HCl remaining / Volume of HCl solution (L)
= 0.00400 moles / 0.03000 L
≈ 0.133 M
Molarity of Ca(OH)2 = Moles of Ca(OH)2 / Volume of Ca(OH)2 solution (L)
= 0.00200 moles / 0.02000 L
= 0.100 M
Molarity of CaCl2 = Moles of CaCl2 formed / Volume of Ca(OH)2 solution (L)
= 0.00200 moles / 0.02000 L
= 0.100 M
Therefore, the molarity of HCl is approximately 0.133 M, the molarity of Ca(OH)2 is 0.100 M, and the molarity of CaCl2 is 0.100 M after the reaction has stopped.