MgCl 2 (aq)+2 NaOH(aq) →Mg(OH) 2 (s)+2 NaCl(aq) f 0.264 MMgC*l_{2} reacts with a 45.31 of 0.759 M to produce Mg(OH) 2 and NaCl calculate the mass of Mg(OH)2 that can be produced

Answer 1

MgCl2(aq) + 2NaOH(aq) ==> Mg(OH)2(s) + 2NaCl(aq) ... balanced equation

Since we are given the amounts of both reactants, the first thing we must do is to determine which reactant is present in limiting supply. One way to do this is to simply divide the moles of each reactant by the corresponding coefficient in the balanced equation. Whichever value is less represents the limiting reactant.

Unfortunately, you did not provide the volume of 0.264 M MgCl2, so we cannot determine the moles of MgCl2 present.

The moles of NaOH present = 45.31 ml x 1 L / 1000 mls x 0.759 mol/L = 0.0344 mols

We would do the same calculation for MgCl2 if we knew the volume (which was omitted).

Please re-submit the question with the volume of MgCl2 that was used.