57.0 mL of 3.40 M sodium hydroxide is combined with 38.0 mL of 1.90 M magnesium chloride. What mass in grams of solid forms?

4.06 g of Mg(OH)2

Explanation:

The equation of the reaction is;

2NaOH(aq) + MgCl2(aq) > Mg(OH)2(s) + 2NaCl(aq)

Number of moles of NaOH= 57/1000 * 3.40 = 0.19 moles

Number of moles of MgCl2 = 38/1000 * 1.9 = 0.07 moles

We must find the limiting reactant. This is the reactant that yields the least amount of the solid.

From the reaction equation;

2 moles of NaOH yields 1 mole of Mg(OH)2

0.19 moles of NaOH yields 0.19 * 1/2 = 0.095 moles of Mg(OH)2

Also;

1 mole of MgCl2 yields 1 mole of Mg(OH)2

0.07 moles of MgCl2 also yields 0.07 moles of Mg(OH)2

Hence MgCl2 is the limiting reactant and 0.07 moles of Mg(OH)2 is produced.

Mass of Mg(OH)2 produced = number of mole * molar mass

molar mass of Mg(OH)2 = 58 g/mol

Mass of Mg(OH)2 produced = 0.07mol * 58 g/mol = 4.06 g of Mg(OH)2

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