What volume of 0.140 M HCl is needed to neutralize 2.53g of Mg(OH)
To determine the volume of 0.140 M HCl needed to neutralize 2.53g of Mg(OH)2, we can use the balanced chemical equation for the reaction between HCl and Mg(OH)2.
The balanced equation is: 2HCl + Mg(OH)2 -> MgCl2 + 2H2O
From the equation, we can see that 2 moles of HCl react with 1 mole of Mg(OH)2. We need to convert the mass of Mg(OH)2 to moles, and then use the mole ratio to calculate the moles of HCl required to neutralize it.
First, calculate the moles of Mg(OH)2:
molar mass of Mg(OH)2 = 24.31 g/mol (Mg) + 16.00 g/mol (O) + 2 * 1.01 g/mol (H)
= 58.33 g/mol
moles of Mg(OH)2 = mass / molar mass
= 2.53 g / 58.33 g/mol
≈ 0.0434 mol
According to the balanced equation, 2 moles of HCl are required for every 1 mole of Mg(OH)2. Therefore, the moles of HCl needed can be calculated as:
moles of HCl = 2 × moles of Mg(OH)2
= 2 × 0.0434 mol
= 0.0868 mol
Now, we can calculate the volume of 0.140 M HCl needed to neutralize this amount of HCl. The equation for calculating the volume is:
volume (L) = moles / concentration (M)
volume = 0.0868 mol / 0.140 M
≈ 0.620 L
To convert the volume from liters to milliliters, we multiply by 1000.
volume = 0.620 L × 1000
= 620 mL
Therefore, approximately 620 mL of 0.140 M HCl is needed to neutralize 2.53 g of Mg(OH)2.