Asked by noorie
10HI+2KMNO4+3H2SO4-5I2+2MNSO4+K2SO4+8H20
Find the mass of water produced when 300g HI reacts with 30 g KMno4 and excess H2So4.
Find the mass of k2So4 produces 30 mL of .250 M KMno4 reacts with 400 mL .100 M HI.
Find the mass of water produced when 300g HI reacts with 30 g KMno4 and excess H2So4.
Find the mass of k2So4 produces 30 mL of .250 M KMno4 reacts with 400 mL .100 M HI.
Answers
Answered by
DrBob222
mols HI = grams/molar mass = ?
mols KMnO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols HI to mols H2O.
Same procedure convert mols KMnO4 to mols H2O.
It is likely that the values for mol;s H2O will be different which means one of them must be wrong. In limiting reagent problems, which this is, the smaller values is ALWAYS the correct one and the reagent producing that number is the limiting reagent.
Using the smaller value, convert to grams H2O. grams = mols H2O x molar mass H2O
mols KMnO4 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols HI to mols H2O.
Same procedure convert mols KMnO4 to mols H2O.
It is likely that the values for mol;s H2O will be different which means one of them must be wrong. In limiting reagent problems, which this is, the smaller values is ALWAYS the correct one and the reagent producing that number is the limiting reagent.
Using the smaller value, convert to grams H2O. grams = mols H2O x molar mass H2O
Answered by
Anonymous
i do not get it why convert grams H2O. grams = mols H2O x molar mass H2O.
Because you already got the smallest value(limiting reagent)
Because you already got the smallest value(limiting reagent)
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