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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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Respuesta :

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