n 7 cf 13 Sapling Learning Chlorine gas can be prepared in t

n 7 cf 13 Sapling Learning Chlorine gas can be prepared in the laboratory by the reaction of hydrochioric acld with manganese(V) oxide You add 36.7 g of MnO2 to a solution containing 41.3 g of HCI (a) What is the limiting reactant? (b) What is the theoretical yield of Ch? g Cl, 0 1 0 2 (c) If the yield of the is 81.3%, what is the actual yield of chlorine? g Cl O Type here to search

Solution

a) moles of MnO2 = mass of MnO2 / molar mass of MnO2

= 36.7 g / 86.93 g mol-1

= 0.422 mol

So, in given reaction moles of MnO2 = 0.422 * 1 = 0.422 mol

Similarly for HCl,

moles of HCl = 41.3 g / 36.46 gmol-1 = 1.132 mol

In given reaction, mol of HCl = 1.132 mol * 4 = 4.53 mol

Since mol of MnO2 is lesser in amount , it is the limiting reagent.

b) theoratical yield = moles of limiting reagent * (stoichiometric coefficient of product / stoichiometric coefficient of Limiting reagent ) * molar mass of product (Cl2)

So, theoretical yield = 0.422 mol * (1 / 1) * 70.90 gmol-1

Theoretical yield = 29.91 g

c) since, % yield = actual yield / theoratical yield

(81.3 / 100) * 29.91 g = actual yield

24.31 g = actual yield