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What mass of precipitate (in g) is formed when 250.0 mL of 0.150 M CuCl₂ is mixed with excess KOH in the following chemical reaction?
CuCl₂(aq) + 2 KOH(aq) → Cu(OH)₂(s) + 2 KCl(aq)

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ANUSARA

Answer:

3.6487g

CuCl2 moles reacted = (0.15×250)/1000

according to balanced chemical equation

precipitated Cu(OH)2 moles = Reacted CuCl2 moles

molar mass of Cu(OH)2 = 63.5+ (17+1)×2 = 97.5

mass of precipitate = (97.5 × 0.15×250)/1000

= 3.648g

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