Asked by Brianna
Consider a 1.50-g mixture of magnesium nitrate and magnesium chloride.After dissolving this mixture in water,0.500M silver nitrate is added dropwise until precipitate formation is complete. THe mass of the white precipitate formed is 0.641 g.
a)Calculate the mass percent of magnesium chloride in the mixture.
b)Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.
a)Calculate the mass percent of magnesium chloride in the mixture.
b)Determine the minimum volume of silver nitrate that must have been added to ensure complete formation of the precipitate.
Answers
Answered by
DrBob222
Convert the mass of the ppt (which is AgCl) to mols. # mols = g/molar mass.
Convert mols AgCl to mols MgCl2.
Convert mols MgCl2 to g MgCl2. g = mols x molar mass.
Convert g MgCl2 to percent.
%MgCl2 = [mass MgCl2/1.5 g]*100 = ??
b.
Look at the definition of molarity.
M = mols/L.
You know M and you know mols. Calculate L and convert to mL if required. Post your work if you get stuck.
Convert mols AgCl to mols MgCl2.
Convert mols MgCl2 to g MgCl2. g = mols x molar mass.
Convert g MgCl2 to percent.
%MgCl2 = [mass MgCl2/1.5 g]*100 = ??
b.
Look at the definition of molarity.
M = mols/L.
You know M and you know mols. Calculate L and convert to mL if required. Post your work if you get stuck.
Answered by
Brianna
i cant figure out my equation.
i have to many.this is waht i got:
MgNO3+MgCl+AgNO3->AgCl+MgCl2+2NO3-
I really don't think that's right?!
i have to many.this is waht i got:
MgNO3+MgCl+AgNO3->AgCl+MgCl2+2NO3-
I really don't think that's right?!
Answered by
Anonymous
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