Chemistry, 29.07.2021 06:30 21hendlill

If 50.0 g of sulfuric acid and 40.0 grams of barium chloride are mixed, how many grams of sulfuric acid and how many grams of barium chloride remain after the double replacement reaction is complete?

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After the double replacement reaction from mixing 50.0g of sulfuric acid and 40.0 grams of barium chloride is complete, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain.

The equation of the reaction between sulfuric acid and barium chloride is

BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

From this equation of reaction, it means 1 mole of barium chloride will completely react with 1 mole of sulfuric acid.

From the question, we have 50.0g of sulfuric acid and 40.0 grams of barium chloride.

First, we will determine the number of moles of the sulfuric acid and barium chloride present.

Number of moles is given by the formula

Number of moles = Mass / Molar mass

For sulfuric acid

Mass = 50.0 g

Molar mass = 98.079 g/mol

∴ Number of moles = 50.0 / 98.079

Numbers of moles of sulfuric = 0.509793 mol

For barium chloride

Mass = 40.0 grams

Molar mass of barium chloride = 208.23 g/mol

∴ Number of moles = 40.0 / 208.23

Number of moles of barium chloride = 0.192095 mol

Since the number of moles of sulfuric acid is more than that of barium chloride, then the limiting reagent is barium chloride and the excess reagent is sulfuric acid

NOTE: A limiting reagent is the reactant that is completely used up in a reaction, and it determines when the reaction stops.

Hence, barium chloride will be used up during the reaction (that is, 0 grams will remain after the reaction is complete).

For the mass of sulfuric acid that will remain,

First, we will determine the number of mole that will remain.

Since 1 mole of barium chloride completely reacts with 1 mole of sulfuric acid, then 0.192095 mol of barium chloride will react with 0.192095 mol of sulfuric acid.

∴ The remaining number moles of sulfuric acid = 0.509793 mol - 0.192095 mol

The remaining number moles of sulfuric acid = 0.317698 mol

Then, from

Mass = Number of moles × Molar mass

Mass = 0.317698 mol ×98.079 g/mol

Mass = 31. 1595 gram

Mass ≅ 31.16 grams

∴ 31.16 grams remains after the reaction is complete.

Hence, 31.16 grams of sulfuric acid and 0 grams of barium chloride remain after the double replacement reaction is complete.

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If 50.0 g of sulfuric acid and 40.0 grams of barium chloride are mixed, how many grams of sulfuric a...

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