Answer to Question #240362 in General Chemistry for sway

Question

For the following reaction, 79.9 grams of barium chloride are allowed to react with 70.8 grams of potassium sulfate.

barium chloride (aq) + potassium sulfate (aq) barium sulfate (s) + potassium chloride (aq)

What is the maximum amount of barium sulfate that can be formed?  grams

What is the FORMULA for the limiting reagent? 

What amount of the excess reagent remains after the reaction is complete?  grams 

Answer 1

BaCl2 + K2SO4 -------- BaSO4 + 2KCl

We need to find the limiting reagent first to know which species will be used to produce max amount of the product.

208g of BaCl2 reacts with 174g of K2SO4

79.9g - - - - - xg of K2SO4

x=66.84g

Therefore BaCl2 will be used

1)If 208g of BaCl2 produces 233g of BaSO4

79.9g will produce xg of BaSO4

x=89.5g

2)The limiting reagent is BaCl2

3)70.8g - 66.84g

=3.96g