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

4 HCl ( aq ) + MnO 2 ( s ) ⟶ MnCl 2 ( aq ) + 2 H 2 O ( l ) + Cl 2 ( g )
A sample of 37.3 g MnO 2 37.3 g MnO2 is added to a solution containing 50.1 g 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. HCl

b. 24.3 g of Cl₂

c. 19.4 g of Cl₂

Explanation:

a. In order to determine the limiting reactant we convert the mass of each reactant to moles.

The reaction is: 4HCl (aq) + MnO₂(s) → MnCl₂(aq) + 2H₂O (l) + Cl₂(g)

37.3 g / 86.94g/mol = 0.429 moles of manganese dioxide

50.1 g / 36.45 g/mol = 1.37 moles of HCl

Ratio is 1:4. 1 mol of MnO₂ needs 4 moles of hydrochloric to react

Then, 0.429 moles of dioxide will react with (0.429 . 4) /1 = 1.71 moles of HCl.

At least, we have 1.37 moles, so the HCl is the limiting reactant.

b. In this question, we can work with stoichiometry with a rule of three:

4 moles of HCl produce 1 mol of chlorine

Then, 1.37 moles of HCl will produce (1.37 . 1) /4 = 0.3425 moles of Cl₂

We convert the moles to mass → 24.3 g of Cl₂

c. In order to determine the actual yield of chlorine, we multiply

Theoretical yield . Yield percent/100 → 24.3 g . 79.9/100 = 19.4 g

calculate how many grams of ammonia can be produced from 4.21 grams of

n2 + 3h2 = 2nh3

my answer:

25.0g n2 x (1 mol/ 28.0gn2) x (2 mol nh3/ 1 mol n2) x (17.0g nh3/ 1 molnh3) = 30.36g n2

25.0g h2 x (1 mol h2/ 2.00g h2) x (2 mol nh3/ 3 mol h2) x (17.0g nh3/ 1 mol nh3) = 141.67g h2

lr = n2