What volume of .500 M HCl is needed to neutralize .287g Mg(OH)2?
To find the volume of HCl needed to neutralize Mg(OH)2, we first need to balance the chemical equation of the reaction between HCl and Mg(OH)2:
2HCl + Mg(OH)2 -> MgCl2 + 2H2O
From the balanced equation, we can see that it takes 2 moles of HCl to react with 1 mole of Mg(OH)2.
Now let's calculate the number of moles of Mg(OH)2:
molar mass of Mg(OH)2 = 24.31 g/mol + 16.00 g/mol + 2*(1.01 g/mol) = 58.33 g/mol
moles of Mg(OH)2 = mass / molar mass = 0.287 g / 58.33 g/mol = 0.00492 mol
Since it takes 2 moles of HCl to react with 1 mole of Mg(OH)2, we need 2 * 0.00492 mol = 0.00984 mol of HCl.
Now, let's calculate the volume of 0.500 M HCl needed:
molarity (M) = moles / volume (L)
0.500 M = 0.00984 mol / volume
volume = 0.00984 mol / 0.500 mol/L = 0.0197 L = 19.7 mL
Therefore, 19.7 mL of 0.500 M HCl is needed to neutralize 0.287 g of Mg(OH)2.
To determine the volume of 0.500 M HCl needed to neutralize 0.287 g of Mg(OH)2, we need to follow these steps:
Step 1: Write the balanced chemical equation for the reaction between HCl and Mg(OH)2.
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
Step 2: Calculate the number of moles of Mg(OH)2.
Given mass of Mg(OH)2 = 0.287 g
Molecular weight of Mg(OH)2 = (1 x Atomic weight of Mg) + (2 x Atomic weight of O) + (2 x Atomic weight of H)
= (1 x 24.31) + (2 x 16.00) + (2 x 1.01)
= 24.31 + 32.00 + 2.02
= 58.33 g/mol
Number of moles of Mg(OH)2 = Given mass / Molecular weight
= 0.287 g / 58.33 g/mol
= 0.00492 mol
Step 3: Use stoichiometry to determine the number of moles of HCl needed.
From the balanced equation, we see that:
1 mol of Mg(OH)2 reacts with 2 mol of HCl
Number of moles of HCl needed = 2 x Number of moles of Mg(OH)2
= 2 x 0.00492 mol
= 0.00984 mol
Step 4: Calculate the volume of 0.500 M HCl needed.
Molarity (M) = Moles / Volume (L)
Rearranging the equation, we have:
Volume (L) = Moles / Molarity
Volume of HCl needed = Number of moles of HCl needed / Molarity of HCl
= 0.00984 mol / 0.500 mol/L
= 0.0197 L
= 19.7 mL
Therefore, you would need 19.7 mL of 0.500 M HCl to neutralize 0.287 g of Mg(OH)2.