How many moles of magnesium chloride are formed when 55 mL of 0.70 M HCl is added to 42 mL of 1.204 g Mg(OH)2?
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
A. 0.019 mol
B. 0.012 mol
C. 0.021 mol
D. 0.039 mol
This is a limiting reagent (LR) problem. You know that when values are given for BOTH reactants.
Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)
mols HCl = M x L = approx 0.038 but you need a better answer than that estimate.
mols Mg(OH)2 = approx 0.05
Convert mols HCl to mols MgCl2 if we had all of the Mg(OH)2 we needed.
That's 0.038 x (1 mol MgCl2/2 mols HCl) = approx 0.038 x 1/2 = approx 0.019
Convert mols Mg(OH)2 to mols MgCl2 if we had all of the HCl needed. That's
0.05 x (1 mol MgCl2/1 mol Mg(OH)2) = 0.0 x 1/1 = approx 0.05
Both of these answers can't be right. In LR problems the correct value is ALWAYS the smaller value and the reagent producing that value is the LR. So HCl is the LR and approx 0.019 mols MgCl2 will be formed. I wouldn't pick A as the answer until going through and getting better numbers than my estimates.
Well, let's calculate it step by step, but don't worry, I'll keep it light and funny!
First, let's convert the volume of the HCl solution from mL to L, because moles are usually in terms of liters. So, 55 mL of 0.70 M HCl is equal to 0.055 L.
Next, to find the moles of magnesium hydroxide (Mg(OH)2), we'll use its molar mass. The molar mass of Mg(OH)2 is 58.33 g/mol, which is definitely not light as a feather!
Now, we can calculate the moles of Mg(OH)2 using the given mass of 1.204 g. It's a simple division, but let's multiply it by the Avogadro's number just to feel fancy. So, 1.204 g * (1 mol/58.33 g) * (6.022 x 10^23 mol^(-1)).
The answer is 1.267 x 10^22 moles, but let's not overwhelm you with such a big number. Let's round it to 1 mole, because jokes are better when they're rounded!
Now, we can find the number of moles of MgCl2 formed by comparing the balanced equation. It's like comparing two clowns, but without the makeup! For every 1 mole of Mg(OH)2, we get 1 mole of MgCl2. So, the answer is 1 mole of MgCl2.
Finally, let's add the moles of MgCl2 from both solutions. Since we had 0.055 L of 0.70 M HCl, we have 0.055 * 0.70 = 0.0385 moles. And for the Mg(OH)2, we had 1 mole. Adding them together, we get 1.0385 moles, but let's round it to 1 mole, because it's a bit easier on the eyes!
So, the answer is D. 0.039 moles, or should I say, 1 mole of laughs!
To find the number of moles of magnesium chloride formed, we need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
First, let's calculate the number of moles of HCl:
Given:
Volume of HCl solution = 55 mL
Concentration of HCl = 0.70 M
To convert mL to L, divide by 1000:
Volume of HCl solution = 55 mL ÷ 1000 = 0.055 L
Now we can calculate the moles of HCl:
Moles of HCl = Volume (L) x Concentration (M)
Moles of HCl = 0.055 L x 0.70 M = 0.0385 mol
Next, let's calculate the number of moles of Mg(OH)2:
Given:
Volume of Mg(OH)2 solution = 42 mL
Mass of Mg(OH)2 = 1.204 g
Molar mass of Mg(OH)2 = 24.31 g/mol + 2(16.00 g/mol) + 2(1.01 g/mol) = 58.33 g/mol
To convert mL to L, divide by 1000:
Volume of Mg(OH)2 solution = 42 mL ÷ 1000 = 0.042 L
Now we can calculate the moles of Mg(OH)2:
Moles of Mg(OH)2 = Mass (g) ÷ Molar mass (g/mol)
Moles of Mg(OH)2 = 1.204 g ÷ 58.33 g/mol ≈ 0.0206 mol
According to the balanced equation, the ratio between Mg(OH)2 and HCl is 1:2. This means that 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
Since the molar ratio between HCl and Mg(OH)2 is 2:1, we can determine the number of moles of HCl required to react with all the Mg(OH)2 by multiplying the moles of Mg(OH)2 by 2:
Moles of HCl required = 0.0206 mol x 2 = 0.0412 mol
Comparing the moles of HCl required (0.0412 mol) with the moles of HCl available (0.0385 mol), we can see that the moles of HCl available are limiting. Therefore, 0.0385 moles of magnesium chloride will be formed.
The correct answer is D. 0.039 mol.
To find the number of moles of magnesium chloride formed, we will use the stoichiometry of the reaction and the given information.
First, let's find the number of moles of HCl and Mg(OH)2:
To find the moles of HCl:
Volume of HCl = 55 mL = 0.055 L
Molarity of HCl = 0.70 M
Moles of HCl = Volume * Molarity = 0.055 L * 0.70 M = 0.0385 mol
Next, let's find the moles of Mg(OH)2:
Volume of Mg(OH)2 = 42 mL = 0.042 L
Mass of Mg(OH)2 = 1.204 g
Molar mass of Mg(OH)2 = 58.32 g/mol (24.31 g/mol for Mg + 2 * 16.00 g/mol for O + 2 * 1.01 g/mol for H)
Moles of Mg(OH)2 = Mass / Molar mass = 1.204 g / 58.32 g/mol = 0.0207 mol
According to the balanced equation, 1 mole of Mg(OH)2 reacts with 2 moles of HCl to form 1 mole of MgCl2. Therefore, the number of moles of MgCl2 formed will be half the number of moles of HCl used.
Moles of MgCl2 = Moles of HCl / 2 = 0.0385 mol / 2 = 0.0193 mol
Now, we need to round the number of moles to the appropriate number of significant figures. The given options are in three significant figures.
Therefore, the closest option to 0.0193 mol is option A: 0.019 mol.
So, the answer is option A: 0.019 mol.