What Volume of 0.575M HCl will neutralize 4.22g of Mg(OH)2? I got the BRE of 2HCl + 1Mg(OH)2 ---> 1MgCl2 + H2O, what do i do from there?
Convert 4.22 g Mg(OH)2 to mols. mols == grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Mg(OH)2 to mols HCl.
M HCl = mols HCl/L HCl.; You know M and mols, solve for L HCl
2 H20, my bad
To determine the volume of 0.575 M HCl required to neutralize 4.22 g of Mg(OH)2, you can use a balanced chemical equation and the concept of stoichiometry.
First, let's take a look at the balanced chemical equation you provided:
2HCl + Mg(OH)2 → MgCl2 + 2H2O
From this equation, we can see that 2 moles of HCl react with 1 mole of Mg(OH)2 to produce 1 mole of MgCl2 and 2 moles of water.
Here's how you can proceed:
Step 1: Calculate the number of moles of Mg(OH)2.
To do this, use the formula:
moles = mass / molar mass
The molar mass of Mg(OH)2 can be calculated by adding the atomic masses of magnesium (Mg), oxygen (O), and hydrogen (H).
Molar mass of Mg(OH)2 = (24.31 g/mol + 2*(1.01 g/mol + 16.00 g/mol))
Step 2: Determine the moles of HCl required to react with Mg(OH)2.
Since the stoichiometric ratio between HCl and Mg(OH)2 is 2:1, the number of moles of HCl is half the number of moles of Mg(OH)2.
Step 3: Calculate the volume of 0.575 M HCl needed.
Use the following equation to relate moles, concentration, and volume:
moles = concentration * volume
Rearrange the equation to solve for volume:
volume = moles / concentration
In this case, the concentration is given as 0.575 M, and the moles of HCl required have been calculated from Step 2.
By calculating the volume using this method, you can determine the volume of 0.575 M HCl required to neutralize 4.22 g of Mg(OH)2.