Question

Limiting agent for Na2SO4 + BaCl2? Theoritical yield?

Part 2: Na2SO4 + BaCl2 Initial mass of Na 50.:\.009 Predicted mass of Baso, based on mass of Na2SO!! Initial mass of BaClz: 1.01g Predicted mass of BaSO, based on mass of BaCl2: Na2SO4 BaCl2 Limiting reagent: Explain why this is the limiting reagent: Theoretical Yield: Actual mass of Baso, obtained: % Yield for the reaction: different from 100%, why might this be? If this is different from 100%, why

Answer 1

Answer:

1) Initial mass of Na₂SO₄ is 1.0 g

Molecular weight of Na₂SO₄ is 142.04 g/mol

Moles of Na₂SO₄ = Mass of Na₂SO₄ / Molecular weight of Na₂SO₄

                                          = 1.0 / 142.04

                            = 0.0070 moles

The overall reaction is

Na₂SO₄   + BaCl₂     à     BaSO₄   + NaCl

If Na₂SO₄ and BaCl₂ reacts in 1:1 ratio then 0.0070 moles of BaSO₄ is formed.

Therefore, mass of BaSO₄ can be calculated by similar formula

Moles of BaSO₄ = Mass of BaSO₄ / Molecular weight of BaSO₄

Molecular weight of BaSO₄ = 233.38 g/mol

Mass of BaSO₄ = Moles x Molecular weight

                          = 0.0070 x 233.38

                          = 1.633 g

2) Initial mass of BaCl₂ is 1.01g

    Molecular weight of BaCl₂ is 208.23 g/mol

   Moles of BaCl₂ = Mass of BaCl₂ / Molecular weight of BaCl₂

                             = 1.01 / 208.23

                             = 0.0048 moles

The overall reaction is

Na₂SO₄   + BaCl₂     à     BaSO₄   + NaCl

If Na₂SO₄ and BaCl₂ reacts in 1:1 ratio then 0.0048 moles of BaSO₄ is formed.

Therefore, mass of BaSO₄ can be calculated by similar formula

Moles of BaSO₄ = Mass of BaSO₄ / Molecular weight of BaSO₄

Molecular weight of BaSO₄ = 233.38 g/mol

Mass of BaSO₄ = Moles x Molecular weight

                        = 0.0048 x 233.38

                          = 1.131 g

3) The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete.

From the given mass of reactant and obtained mass we have 0.0070 moles of Na₂SO₄ and 0.0048 moles of BaCl₂. The reaction proceed in 1:1 ratio, then 0.0048 moles of BaCl₂ react only with 0.0048 moles of Na₂SO₄. (Leaving behind 0.0022 moles of Na₂SO₄, thus making it excess reagent). Therefore, BaCl₂ is the limiting reagent of reaction.

4) Theoretical Yield can be calculated as

Theoretical Yield =      x weight of limiting reactant

Molecular weight of Product Molecular weight of limiting reactant

         

                          =         x 1.01

233.38 208.23

                          = 1.131 g

5) % yield =     x 100

Actual Yield (actual mass of product) Theoretical Yield

(As actual mass obtained is not given. You can calculate by using the above formula).