When potassium dichromate (K2CrO4) is added to a solution containing 0.500g silver nitrate (AgNO3) solid silver chromate (Ag2CrO4) is formed
The balanced equation is below:
K2CrO4(aq) + 2AgNO3(aq) --> Ag2CrO4(s) + 2KNO3(aq)
It is unclear what is required in this problem, but the mass of AgNO3 allows to determine the mass of the precipitate mentioned in the question.
Molar masses are:
AgNO3 - 169.9 g/mol
Ag2CrO4 - 331.7 g/mol
"m(Ag_2CrO_4)=0.500\\ g(AgNO_3)\\times\\frac{1\\ mol(AgNO_3)}{169.9\\ g(AgNO_3)}\\times\\frac{1\\ mol(Ag_2CrO_4)}{2\\ mol(AgNO_3)}\\times\\frac{331.7\\ g(Ag_2CrO_4)}{1\\ mol(Ag_2CrO_4)}=0.488\\ g"
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