How many grams of Mg(OH)2 are needed to neutralize 55 ml of 0.25 M HCl .

according to the following reaction. (show all calculations for credit)
Mg(OH)2 (aq) + 2HCl(aq) --- MgCl2 (aq) + 2H2O(l)

mols HCl = M x L = ?

Convert mols HCl to mols Mg(OH)2 using the coefficients in the balanced equation.
Convert mols Mg(OH)2 to grams. g = mols x molar mass.

Got the first part. Thanks.

I'm still not sure how to convert mols using the balanced equation...
sorry, this class got hard really fast.

To determine the number of grams of Mg(OH)2 needed to neutralize 55 ml of 0.25 M HCl, we can use the molarity and stoichiometric ratio of the reaction.

First, let's calculate the number of moles of HCl present in the 55 ml of 0.25 M HCl:

Molarity (M) = moles/volume (L)

0.25 M = moles/0.055 L

moles = 0.025 moles

According to the balanced equation, the stoichiometric 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.

Therefore, the number of moles of Mg(OH)2 required to neutralize 0.025 moles of HCl is:

moles of Mg(OH)2 = (0.025 moles HCl) / (2 moles HCl/1 mole Mg(OH)2)

moles of Mg(OH)2 = 0.0125 moles

Finally, we need to convert moles of Mg(OH)2 to grams using the molar mass of Mg(OH)2.

The molar mass of Mg(OH)2 is calculated by adding up the atomic masses of each constituent element:

Mg: 24.31 g/mol
O: 16.00 g/mol
H: 1.01 g/mol (2 hydrogens in Mg(OH)2)

Molar mass of Mg(OH)2 = (24.31 g/mol) + (16.00 g/mol) + (1.01 g/mol * 2) = 58.33 g/mol

Finally, the number of grams of Mg(OH)2 required is:

grams of Mg(OH)2 = moles of Mg(OH)2 * molar mass of Mg(OH)2

grams of Mg(OH)2 = 0.0125 moles * 58.33 g/mol ≈ 0.73 grams

Therefore, approximately 0.73 grams of Mg(OH)2 are needed to neutralize 55 ml of 0.25 M HCl.

To determine the number of grams of Mg(OH)2 needed to neutralize 55 ml of 0.25 M HCl, we can use the concept of stoichiometry.

First, let's calculate the number of moles of HCl in 55 ml of 0.25 M HCl.

Molarity (M) = moles/volume (in liters)
0.25 M = moles/0.055 L (since we have to convert ml to liters)
moles = 0.25 M * 0.055 L
moles = 0.01375

From the balanced chemical equation, we can see that 1 mole of Mg(OH)2 reacts with 2 moles of HCl.

Therefore, to neutralize the HCl, we need half the amount of moles of Mg(OH)2. So, we divide the moles of HCl by 2.

Moles of Mg(OH)2 = 0.01375 / 2
Moles of Mg(OH)2 = 0.006875

Lastly, we need to convert moles of Mg(OH)2 to grams.

To do this, we need the molar mass of Mg(OH)2, which is:
(24.31 g/mol) + 2(16.00 g/mol) + 2(1.01 g/mol) = 58.33 g/mol

Grams of Mg(OH)2 = moles * molar mass
Grams of Mg(OH)2 = 0.006875 * 58.33
Grams of Mg(OH)2 = 0.4007 g

Therefore, approximately 0.4007 grams of Mg(OH)2 are needed to neutralize 55 ml of 0.25 M HCl.