Question

10)Aluminum bromide and potassium sulfate react to form potassium bromide and aluminum sulfate.

a. Write and balance the reaction

b. If you have 250.0 g of aluminum bromide and 200.0 g of potassium sulfate, which reagent is limiting?

c. What would be the theoretical yield of potassium bromide?

d. How much of the excess reactant is left over?

Answer 1

the balanced equation is as follows

2 AlBr3 + 3 K2SO4 ----------------------> Al2(SO4)3 + 6 KBr

moles = mass / molar mass

for AlBr3 => 250.0g / 266.69 g/mol => 0.9374 moles

for K2SO4 => 200 g / 174.259 g/mol => 1.1477 moles

2mol AlBr3 -----------------> 6 mol KBr

0.9374 mol -------------------> ?

=> 0.9374 * 6 / 2 => 2.81225 moles

3 moles K2SO4 ----------------------> 6 mol KBr

1.1477 mol -------------------------> ?

=>1.1477 * 6 / 3 => 2.2954 moles

limiting reactant is K2SO4

mass of KBr ( theoretical yield) formed => 2.2954 moles * 119.002 g/mol => 273.16 grams

excess reactant moles and mass => 0.9374 - 0.765 => 0.17227 moles

mass is 0.17227 * 266.69 => 45.943 grams