Answer to Question #314629 in General Chemistry for Ivan Ackerman

Question

The reaction goes to completion, but in the process of washing and drying the precipitate, some was lost. The percent yield for the reaction is 79.9%

79.9%. How many grams of precipitate are recovered?

precipitate recovered:

How many grams of the excess reactant remain? Assume the reaction goes to completion.

excess reactant remaining:

Answer 1

2KClO3→2KCl+3O2

From the reaction,it can be seen that;

2mol of KClO3= 3mol of O2

1 mol of KClO3=3/2 mol O2

=1.5 mol O2

The yield of the reaction is given in % i.e 79.9%

1 mil KClO3 generates 1.5 mol O2

Since the yield is 79.9%,the number of moles of O2 generated per mole of KClO3 is

1.5mol×79.9/100=1.1985mol

Number of moles ofO2=42.0g/32.0g

=1.31 mol.

Number of moles of KClO3

=1.31mol O2/1.1985 mol O2/KClO3

=1.093 mol KClO3

Weight of KClO3=1.35 mol KClO3×molecular weight of KClO3

=1.093×112.6g

=123.1g