what volume of 1.015mol/L magnesium hydroxide is needed to neutralize 40.0mL of 1.60mol/L hydrochloric acid.
Can you work it out please? Thanks!
Of course! To solve this problem, we can use the concept of stoichiometry and the balanced chemical equation between magnesium hydroxide (Mg(OH)2) and hydrochloric acid (HCl). The balanced equation for the reaction is:
Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
The stoichiometry of the equation tells us that 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
To find the volume of Mg(OH)2 needed, we will follow these steps:
Step 1: Calculate the number of moles of HCl in 40.0 mL of 1.60 mol/L HCl.
Number of moles of HCl = (concentration of HCl) x (volume of HCl) = (1.60 mol/L) x (0.0400 L) = 0.0640 mol
Step 2: Using the stoichiometry of the balanced equation, calculate the number of moles of Mg(OH)2 that will react with the calculated moles of HCl.
From the balanced equation, 1 mole of Mg(OH)2 reacts with 2 moles of HCl.
Therefore, the number of moles of Mg(OH)2 = (0.0640 mol HCl) / (2 mol HCl/mol Mg(OH)2) = 0.0320 mol
Step 3: Calculate the volume of 1.015 mol/L Mg(OH)2 needed to supply 0.0320 mol.
Volume of Mg(OH)2 = (number of moles of Mg(OH)2) / (concentration of Mg(OH)2)
Volume of Mg(OH)2 = 0.0320 mol / (1.015 mol/L) = 0.0316 L or 31.6 mL (rounded to three decimal places).
So, to neutralize 40.0 mL of 1.60 mol/L HCl, you will need approximately 31.6 mL of 1.015 mol/L Mg(OH)2.
Write the equation and balance it.
Mg(OH)2 + 2HCl ==> MgCl2 + 2H2O
moles HCl = M x L = ??
moles Mg(OH)2 (look at the coefficients in the balanced equation) needed = 1/2 moles HCl
M HCl = moles HCl/L HCl
Solve for L and convert to mL.