[A table of molecular weights is given below.] How many grams of MnF3 can be produced...
[A table of molecular weights is given below.]
How many grams of MnF3 can be produced by the reaction
of 56.00 grams of MnI2 and 38.00 grams of F2
according to the equation below?
2 MnI2(s) + 13 F2(g) → 2
MnF3(s) + 4 IF5(l)
Answer grams of MnF3
When the reaction is complete it is found that only 11.69 grams of
MnF3 were produced? Determine the %Yield for this
reaction.
Answer % Yield
Molecular Weights
MnI2 308.738 g/mol
F2 37.996 g/mol
MnF3 111.932 g/mol
IF5 221.89 g/mol
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[A table of molecular weights is given below.]
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