## Answers

**Answer:**

**1)** Initial mass of Na_{2}SO_{4} is 1.0 g

Molecular weight of Na_{2}SO_{4} is 142.04 g/mol

Moles of Na_{2}SO_{4} = Mass of Na_{2}SO_{4} / Molecular weight of Na_{2}SO_{4}

_{ } = 1.0 / 142.04

= 0.0070 moles

The overall reaction is

Na_{2}SO_{4} + BaCl_{2} à BaSO_{4} + NaCl

If Na_{2}SO_{4} and BaCl_{2} reacts in 1:1 ratio then 0.0070 moles of BaSO_{4} is formed.

Therefore, mass of BaSO_{4} can be calculated by similar formula

Moles of BaSO_{4} = Mass of BaSO_{4} / Molecular weight of BaSO_{4}

Molecular weight of BaSO_{4} = 233.38 g/mol

**Mass of BaSO _{4} = Moles x Molecular weight**

** = 0.0070 x 233.38**

** = 1.633 g**

**2)** Initial mass of BaCl_{2} is 1.01g

Molecular weight of BaCl_{2} is 208.23 g/mol

Moles of BaCl_{2} = Mass of BaCl_{2} / Molecular weight of BaCl_{2}

= 1.01 / 208.23

= 0.0048 moles

The overall reaction is

Na_{2}SO_{4} + BaCl_{2} à BaSO_{4} + NaCl

If Na_{2}SO_{4} and BaCl_{2} reacts in 1:1 ratio then 0.0048 moles of BaSO_{4} is formed.

Therefore, mass of BaSO_{4} can be calculated by similar formula

Moles of BaSO_{4} = Mass of BaSO_{4} / Molecular weight of BaSO_{4}

Molecular weight of BaSO_{4} = 233.38 g/mol

**Mass of BaSO _{4} = 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 Na_{2}SO_{4} and 0.0048 moles of BaCl_{2}. The reaction proceed in 1:1 ratio, then 0.0048 moles of BaCl_{2} react only with 0.0048 moles of Na_{2}SO_{4}. (Leaving behind 0.0022 moles of Na_{2}SO_{4}, thus making it excess reagent). **Therefore, BaCl _{2} is the limiting reagent of reaction**.

**4)** Theoretical Yield can be calculated as

Theoretical Yield = x weight of limiting reactant

= * * x 1.01

= 1.131 g

**5)** % yield = x 100

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