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 reac
tion?
CuCl₂(aq) + 2 KOH(aq) → Cu(OH)₂(s) + 2 KCl(aq)
1 answer:
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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La respuestas de lo que descirbiste es 18.02 gramos
Saludos
Moles=volume*concentration
=0.1*.83
=.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2
Mass =Moles*Rfm of CO2
=.083*(12+16+16)
=3.7grams
