For the following reaction, 67.1 grams of barium hydroxide are allowed to react with 36.3 grams of sulfuric acid. barium hydroxide (aq) sulfuric acid (aq) barium sulfate (s) water (l) What is the maximum amount of barium sulfate that can be formed

86.2 g of will be produced from the given masses of both reactants.

Explanation:

To calculate the moles, we use the equation:

a) moles of

b) moles of

According to stoichiometry :

1 moles of require 1 mole of

Thus 0.37 moles of   require=  of

Thus is the limiting reagent as it limits the formation of product.

As 1 mole of give =  1 mole of [texBaSO_4[/tex]

Thus 0.37 moles of give =  of

Mass of

Thus 86.2 g of will be produced from the given masses of both reactants.

2b2t

explanation:

2b2t