What volume of 0.140 M HCl is needed to neutralize 2.53g of Mg(OH)_2
Mg(OH)2 + 2HCl ==> 2H2O + MgCl2
mols Mg(OH)2 = grams/molar mass
Convert mols Mg(OH)2 to mols HCl using the coefficients in the balanced equation. That's mols HCl = 2x mols Mg(OH)2.
Then M HCl = mols HCl/L HCl. You know mols HCl and M HCl, substitute and solve for L HCl.
To solve this problem, we can use the concept of stoichiometry and the balanced chemical equation between HCl and Mg(OH)2 to determine the volume of HCl needed to neutralize the given mass of Mg(OH)2.
1. Write the balanced chemical equation:
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
2. Calculate the molar mass of Mg(OH)2:
Mg: 24.31 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Molar mass = (24.31 g/mol × 1) + (16.00 g/mol × 2) + (1.01 g/mol × 2)
= 58.33 g/mol
3. Calculate the number of moles of Mg(OH)2:
Moles = mass / molar mass
Moles = 2.53 g / 58.33 g/mol
≈ 0.0434 mol
4. According to the balanced equation, 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
5. Convert moles of Mg(OH)2 to moles of HCl:
Moles of HCl = 2 × moles of Mg(OH)2
= 2 × 0.0434 mol
≈ 0.0868 mol
6. Finally, calculate the volume of HCl:
Volume = moles / molarity
Volume = 0.0868 mol / 0.140 mol/L
≈ 0.620 L
Therefore, the volume of 0.140 M HCl needed to neutralize 2.53g of Mg(OH)2 is approximately 0.620 L.
To find the volume of 0.140 M HCl needed to neutralize 2.53g of Mg(OH)2, we can use the concept of stoichiometry and the balanced chemical equation.
The balanced chemical equation between HCl and Mg(OH)2 is:
2HCl + Mg(OH)2 -> MgCl2 + 2H2O
From the stoichiometry of the equation, we can see that it takes 2 moles of HCl to react with 1 mole of Mg(OH)2.
First, we need to convert the mass of Mg(OH)2 into moles. The molar mass of Mg(OH)2 is:
Mg: 24.31 g/mol
O: 16.00 g/mol
H: 1.01 g/mol
Molar mass of Mg(OH)2 = 24.31 + (16.00 * 2) + (1.01 * 2)
= 58.33 g/mol
Number of moles of Mg(OH)2 = Mass / Molar mass
= 2.53g / 58.33 g/mol
= 0.0434 mol
Since the stoichiometry tells us that it takes 2 moles of HCl to neutralize 1 mole of Mg(OH)2, we can say that:
0.0434 mol Mg(OH)2 * 2 mol HCl / 1 mol Mg(OH)2 = 0.0868 mol HCl
Now, we can use the concentration and the moles of HCl to find the volume of HCl needed.
The equation we can use is:
Volume (L) = Moles / Concentration (M)
Volume of HCl = 0.0868 mol / 0.140 M
= 0.620 L
Therefore, the volume of 0.140 M HCl needed to neutralize 2.53g of Mg(OH)2 is 0.620 liters.