Question

What volume of 0.244 M NaOH is required to react with 50.0 mL of 0.329 M MgCL₂?

MgCL_2(aq) + NaOH_(aq) = NaCL_(aq) + Mg(OH)_2(aq)

(NOT BALANCED)

Answer 1

BALANCED EQUATION:

MgCl₂(aq) + 2NaOH(aq) = 2NaCl(aq) + Mg(OH)₂(aq)

Here, 1 mole of MgCl₂ reacts with 2 moles of NaOH.

Now, 0.329 M MgCl₂,

0.329 moles of MgCl₂ in 1000 mL of solvent

0.01645 moles of MgCl₂ in 50 mL of solvent.

So, 0.01645 moles of MgCl₂ will react with (2 x 0.01645 =0.0329) moles of MgCl₂.

NaOH is 0.244 M

0.224 moles of NaOH in 1000 mL of solvent.

0.0329 moles of NaOH will be in (1000 x 0.0329)/0.224 mL of solvent.

0.0329 moles of NaOH will be in 146.875 mL of solvent.

Hence 146.875 mL of 0.244 M NaOH is required to react with 50.0 mL of 0.329 M MgCl₂.