Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide:

4HCL(aq) + MnO2(s) ----> MnCl2(aq) + 2H2O(l) + Cl2(g)
You add 42.5 g of MnO2 to a solution containing 47.7 g of HCl.
(a) What is the limiting reactant? MnO2 or HCL?
(b)What is the theortical yield of CO2?
(c) If the yield of the reaction is 79.9%, what is the actual yield of chlorine?

Answer 1

(a) HCl is the limiting reactant.

(b) 28.45 g

(c) 19.10 g.

Step-by-step explanation:

(a) 4HCL(aq) + MnO2(s) ----> MnCl2(aq) + 2H2O(l) + Cl2(g)

Using the relative atomic masses:

4 * (1.008 + 35.45( g of HCl react with (54.983 + 2*15.999) g of MnO2.

145.832 g HCl reacts with 86.981 g MnO2.

So 47.7 g HCl reacts with (86.981 / 145.832) * 47.7 = 28.451 g MnO2.

We have 42.5 g of MnO2 so HCl is the limiting reactant.

(b) The theoretical yield of Cl2 is (70.9 /86.981) * 28.451 = 23.91 g.

(c) The actual yield of chlorine is 23.91 * 0.799 = 19.10 g.

Answer 2

(a) Limiting reactant = HCl

(b) 0.33 moles Cl₂

(c) 0.26 moles Cl₂

Step-by-step explanation:

The equation is:

4HCl(aq) + MnO₂(s) → MnCl₂(aq) + 2H₂O(l) + Cl₂(g) (1)

We have:

m MnO₂ = 42.5 g

m HCl = 47.7 g

M HCl: molar mass = 36.46 g/mol

M MnO₂ = 86.94 g/mol

The number of moles (η) of HCl and MnO₂ is:

`\eta_(HCl) = (m)/(M) = (47.7 g)/(36.46 g/mol) = 1.31 mol`

`\eta_{MnO_(2)} = (42.5 g)/(86.94 g/mol) = 0.49 mol`

(a) From equation (1) we have that 4 moles of HCl reacts with 1 mol of MnO₂, so the limiting reactant is:

`(1 mol MnO_(2))/(4 moles HCl) \cdot 1.31 moles HCl = 0.33 moles MnO_(2)`

We can see that we need 0.33 moles of MnO₂ to react with HCl, but we have 0.49 moles in total, hence, the MnO₂ is in excess and the limiting reactant is HCl.

(b) The theoretical yield of Cl₂ (not CO₂) is the following:

From equation (1) we have that 4 moles of HCl produces 1 mol of Cl₂, hence the moles of Cl₂ produced is:

`\eta_{Cl_(2)} = (1 mol Cl_(2))/(4 moles HCl) \cdot 1.31 moles HCl = 0.33 moles Cl_(2)`

Hence, the theoretical yield of Cl₂ is 0.33 moles.

(c) The actual yield of Cl₂ is:

% = (actual yield*100)/(theoretical yield)

actual yield = 0.799*0.33 moles = 0.26 moles Cl₂.

Therefore, the actual yield of Cl₂ is 0.26 moles.

I hope it helps you!