For the following reaction, 56.6 grams of barium hydroxide are allowed to react with 30.1 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? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Sulfuric acid is limiting reagent.  The formula of limiting reagent is .

Mass of barium sulfate formed is 71.62 g

Excess Mass of barium sulfate remained is 4 g

Explanation:

The formula for the calculation of moles is shown below:

For barium hydroxide

Mass of barium hydroxide = 56.6 g

Molar mass of barium hydroxide = 171.34 g/mol

The formula for the calculation of moles is shown below:

Thus,

For sulfuric acid

Mass of sulfuric acid = 30.1 g

Molar mass of sulfuric acid = 98.079 g/mol

The formula for the calculation of moles is shown below:

Thus,

According to the given reaction:

1 mole of sulfuric acid reacts with 1 mole of barium hydroxide

So,  

0.3069 mole of sulfuric acid reacts with 0.3069 moles of barium hydroxide

Moles of barium hydroxide = 0.3069 moles

Available moles of barium hydroxide = 0.3303 moles

Limiting reagent is the one which is present in small amount. Thus, sulfuric acid is limiting reagent.  The formula of limiting reagent is .

The formation of the product is governed by the limiting reagent. So,

1 mole of sulfuric acid produces 1 mole of barium sulfate

So,  

0.3069 mole of sulfuric acid produces 0.3069 mole of barium sulfate

Moles of barium sulfate = 0.3069 mole

Molar mass of barium sulfate = 233.38 g/mol

Mass of barium sulfate = Moles × Molar mass = 0.3069 × 233.38 g = 71.62 g

Excess reactant = barium hydroxide = 0.3303 moles  - 0.3069 moles   = 0.0234 moles

Molar mass of barium hydroxide = 171.34 g/mol

Mass of barium sulfate = Moles × Molar mass = 0.0234 × 171.34 g = 4 g

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