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?

(a) Limiting reactant = HCl

(b) 0.33 moles Cl₂

(c) 0.26 moles Cl₂

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:

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

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:

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!