Question

If one of my chem students ran the reaction, as stated in problem below, and got 7.052 g of Ba(OH)₂, what is the percent yield of the reaction?

iron (III) sulfate and barium hydroxide, Fe₂(SO₄)₃ + 3Ba(OH)₂ --> 3BaSO₄ + 2Fe(OH)₃

_{If 20.0 g of Fe2(SO4)3 is mixed with 20.0 g of Ba(OH)2.}

_{Molar masses: Fe2(SO4)3 =399.91 Ba(OH)2= 171.35 Fe(OH)3 = 106.88}

_{How many moles of Fe(OH)3 will be produced if all 20.0 g of Fe2(SO4)3 were consumed?}

Answer 1

The balance reaction of iron (III) sulfate and barium hydroxide,asa follows:

Fe₂(SO₄)₃ + 3Ba(OH)₂ --> 3BaSO₄ + 2Fe(OH)₃

Given that 20.0 g of Fe2(SO4)3 and 20.0 g of Ba(OH)2

Number of moles = amount in g / molar mass

Fe2(SO4)3= 20.0 g of Fe2(SO4)3/399.91 g/ mole

= 0.05 moles Fe2(SO4)3

Ba(OH)2.=20.0 g Ba(OH)2 /171.35 g/ mole

= 0.117 moles Ba(OH)2

Ba(OH)2is limiting agent

The limiting agent has following properties:

1. It completely reacted in the reaction.
2. It determines the amount of the product in mole.

Number of mole of Fe(OH)₃ calculated as follows:

0.117 moles Ba(OH)2* 2 mole Fe(OH)₃ /3 mole Ba(OH)2

=0.078 moles Fe(OH)₃

If one of my chem students ran the reaction, as stated in problem below, and got 7.052 g of Ba(OH)₂, thus used barium hydroxid e= 20.0g -4.052 g

= 12.948 g

Percentage yield = observed yield / theoretical yield *100

=12.948/20.0*100

= 64.74%