What mass of precipitate will be produced from the reaction of 50.0 mL of 2.50 mol/L sodium hydroxide with an excess of zinc chloride solution.

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2NaOH + ZnCl2 → 2NaCl + Zn (OH) 2

Zn (OH) 2 is your precipitate.

No. of moles of NaOH = 2.50X50.0:1000 = 0.125mol

2mol of NaOH reacted to produce 1 mol of Zn (OH) 2.

No. of moles of Zn (OH) 2 produced when 0.125mol of NaOH reacted

= 0.125 x 1 : 2

= 0.0625mol

Mass of Zn (OH) 2 = 0.0625 x [65.4+2 (16+1) ] = 6.21g