Question

Help solving table III-E !!

Table III-E. Percent Yield and Amount of Excess Reactant Left Over Mass of strontium sulfate collected (actual yield) 0.39149 Theoretical yield of strontium sulfate (from Table WCC) TID Percent yield of strontium sulfate Mass of excess reactant left over

values for calculations

lab manual explanation

0.394g

Answer 1

This question easily solved by following table|||-E

1-refering data in table |||-B

Actual yield of precipitate collected=(mass filter paper+dried stronium sulfate) -(mass empty filter paper)

=(1.028-.613)gm

=0.415gram

2-since theoretical yield of stronium sulfate is the amount that can be produced from the limiting reactant and in table |||-D Al2(SO4)3 is limiting reactant because it gives less amount of Sr(SO)4.

So theoretical yield of stronium sulfate is 0.416gram

3-percent yield of strontium sulfate={actual yield of sr(So)4}\ {theoretical yield of Sr(So)4}*100

=(0.415gram/0.416gram)*100

=99.75%

4-gram of excess reactant used up= theoretical yeild of sr(So)4=0.416gram

Note-i used this value because i prefer table|||-D previously according to the point 2 in table |||-E.

5-left over grams of excess reactant =initial grams -grams used up or gram of excess reactant used up

=(0.596+0.258)-0.416 [initial gram from table|||-C, sum of mass of reactant ]

=(0.854-0.416)gm

=0.438gm

Table |||-E

Mass of Sr(So)4 collected (actual yeild) =0.415gram

Theoretical yield of strontium sulfate from table|||-D=0.416gram

Percent yield of strontium sulfate =99.75%

Mass of excess reactant left over=0.438gram