1 answer

Limiting agent for Na2SO4 + BaCl2? Theoritical yield? Part 2: Na2SO4 + BaCl2 Initial mass of...

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
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

Answers

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 =  Molecular weight of Product Molecular weight of limiting reactant    x weight of limiting reactant

         

                          =     233.38 208.23    x 1.01

                          = 1.131 g

5) % yield = Actual Yield (actual mass of product) Theoretical Yield     x 100

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

.

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