Question

Consider the following reaction. MgCl2(aq)+2NaOH(aq)⟶Mg(OH)2(s)+2NaCl(aq) A 187.0 mL solution of 0.103 M MgCl2 reacts with a 33.58 mL solution of 0.839 M NaOH to produce Mg(OH)2 and NaCl. Identify the limiting reactant.

Answer 1

Answer:

NaOH is the limiting reactant and MgCl2 is in excess.

Explanation:

let's calculate the mole of each reactant.

number of moles of MgCl2 = $\frac{187}{1000} L* 0.103 mol / L =0.01926 mol$

number of moles of NaOH = $\frac{33.58 }{1000}L * 0.839 mol/L =0.0282 mol$

as per the reaction , 1 mol of Magnesium choloride reacts with 2 moles of NaOH

hence 0.01926 needs 0.01926x2= 0.03852 moles of NaOH , but there are only 0.0282 moles of NaOH available .