Question

1) A student mixes 150. mL of 0.20 M Na₂CO₃ with 130. mL of 0.25 M CaCl₂ to obtain CaCO₃ precipitate.

Which is the limiting reactant?

Select one:

a. sodium carbonate

b. calcium chloride

2) What mass of CaCO₃ should be produced?

Report your answer to the proper significant digits and include units.

ii) The student was able to obtain 2.3 g of precipitate. Calculate the percent yield for the reaction.

Report your answer to the proper significant digits and include units.

Answer 1

Answer -

Given,

Volume of Na₂CO₃ = 150 mL or 0.15 L [1 mL = 0.001 L]

Molarity of Na₂CO₃ = 0.20 M

Volume of CaCl₂ = 130 mL or 0.13 L [1 mL = 0.001 L]

Molarity of CaCl₂ = 0.25 M

Molar Mass of CaCO₃ =100.0869 g/mol

1)

Limiting Reactant = ?

We know that,

Molarity = Moles / Volume in L

So,

Moles = Molarity * Volume in L

So,

Moles of Na₂CO₃ = 0.20 M * 0.15 L

Moles of Na₂CO₃ = 0.03 mol

Moles of CaCl₂ = 0.25 M * 0.13 L

Moles of CaCl₂ = 0.0325 mol

The reaction is,

Na₂CO₃ (aq) + CaCl₂ (aq) → 2 NaCl (aq) + CaCO₃ (s) [BALANCED]

To Find the LIMITING REACTANT,

Divide the available moles by their stiochiometric coefficients, the one which is smaller is the LIMITING REACTANT.

For Na₂CO₃ = 0.03 mol / 1 = 0.03 mol

For CaCl₂ = 0.0325 mol / 1 = 0.0325 mol

So, Na₂CO₃ is the LIMITING REACTANT.

OPTION A is CORRECT. [ANSWER]

2)

Mass of CaCO₃ produced = ?

Na₂CO₃ (aq) + CaCl₂ (aq) → 2 NaCl (aq) + CaCO₃ (s) [BALANCED]

Using Stiochiometry, it can be analyzed that for 1 mole of Na₂CO₃, 1 mole of CaCO₃ is produced.

i.e.

Moles of CaCO₃ produced = Moles of Na₂CO₃

Moles of CaCO₃ produced = 0.03 mol

Now,

Mass = Moles * Molar Mass

So,

Mass of CaCO₃ produced = 0.03 mol * 100.0869 g/mol

Mass of CaCO₃ produced = 3.00 g [ANSWER]

ii)

Actual Yield = 2.3 g

Percentage Yield = ?

We know that,

Percentage Yield = (Actual Yield / Theoretical Yield) * 100

Percentage Yield = (2.3 g / 3.00 g) * 100

Percentage Yield = 76.67 % [ANSWER]