Answer:
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
