Question

2. Write the balanced equation with physical states

3. Calculate maximum/theoretical yield of MgCO3 from each reactant and determine limiting reactant of this reaction. How much excess reactant is there?

Precipitate: 2. Write the balanced equation for the reaction to be performed in this experiment. Be sure to include physical states. CaCl2.2H2O + Na2CO3 3. In another experiment, a student used 0.50 g of MgSO4 and 0.30 g of Na2CO3. a. Calculate the maximum/theoretical yield of MgCO3 from each of these reactants. b. Determine the limiting reactant of this reaction. c. How much excess reactant is there?

Answer 1

2)

In this reaction CaCO₃ will be formed as precipitate. The reaction is a double displacement reaction with precipitation.

$CaCl_2\cdot2H_2O(aq)+Na_2CO_3(aq)\rightarrow CaCO_3(s)+2NaCl(aq)+2H_2O(l)$

3)

(a) and (b)

The reaction between MgSO₄ and Na₂CO₃ will be-

$MgSO_4(aq)+Na_2CO_3(aq)\rightarrow MgCO_3(s)+Na_2SO_4(aq)$

Mass of MgSO₄ = 0.50g

Molar Mass of MgSO₄ = 120.366 g/mol

Moles of MgSO₄ = Mass/ Molar Mass = 0.50 g / (120.366 g/mol) = 4.154 x 10^-3 mol

Mass of Na₂CO₃ = 0.30g

Molar Mass of Na₂CO₃ = 105.989 g/mol

Moles of Na₂CO₃= Mass/ Molar Mass = 0.30 g / (105.989 g/mol) = 2.830 x 10^-3 mol

From the reaction, it is clear that Na₂CO₃ and MgSO₄ reacts in the ratio 1:1.

Thus, because moles of Na₂CO₃ is lesser, it is the limiting reagent

Each mole of Na₂CO₃ gives 1 mole of MgCO₃.

Therefore Moles of MgCO₃ produced = Moles of Na₂CO₃= 2.830 x 10^-3 mol

Molar Mass of MgCO₃ = 81.314 g/mol

Thus, theoretical yield of MgCO₃ = Moles * Molar Mass = 2.830 x 10^-3 mol * 81.314 g/mol = 0.23 g

Hence, theoretical yield of MgCO₃ is 0.23 g

(c)

Excess reactant is MgSO₄

Moles of Excess reactant utilized = 2.830 x 10^-3 mol

Moles of excess reactant left= Initial moles - Moles utilized = 4.154 x 10^-3 mol - 2.830 x 10^-3 mol = 1.324 x 10^-3 mol

Mass of excess ractant = Moles left * Molar Mass = (1.324 x 10^-3 mol)* 120.366 g/mol =0.16 g

Thus, excess reactant left is 0.16 g