How many grams of water are formed when 250 mL of 2.0 M Mg(OH)2 are neutralized by a solution of HCL ?
the ans is 18gm
I agree with ola.
To determine the number of grams of water formed when 250 mL of 2.0 M Mg(OH)2 is neutralized by a solution of HCl, you need to consider the balanced equation for the reaction.
The balanced equation for the neutralization reaction of Mg(OH)2 and HCl is as follows:
Mg(OH)2 + 2HCl → MgCl2 + 2H2O
From the balanced equation, you can see that 1 mol of Mg(OH)2 reacts with 2 moles of HCl, producing 2 moles of water.
Now, let's calculate the number of moles of Mg(OH)2 in 250 mL (0.250 L) of a 2.0 M solution:
mol = Molarity × Volume (L)
mol = 2.0 M × 0.250 L
mol = 0.5 moles of Mg(OH)2
Since 1 mole of Mg(OH)2 produces 2 moles of water, we can determine the number of moles of water formed:
moles of water = 2 moles of Mg(OH)2
Therefore, the number of moles of water formed is 2 times the number of moles of Mg(OH)2, which is 2 × 0.5 moles = 1 mole.
To calculate the mass of water formed, you need to use the molar mass of water, which is approximately 18.015 g/mol.
mass = moles × molar mass
mass = 1 mole × 18.015 g/mol
mass = 18.015 grams of water
Therefore, when 250 mL of 2.0 M Mg(OH)2 is neutralized by a solution of HCl, approximately 18.015 grams of water are formed.