Question

Given that 16.0 g of MnO2 and 30.0 g of HCl react according to: MnO2 + 4HCl--->MnCl2 + Cl2 + 2H2O

a. What is the limiting reagent

b. What is the mass of MnCl2 could be produced?

c. How much of each reagent remains when the reaction is complete?

d. Assuming a % yield of 71.2% calculate the actual yield of MnCl2

Answer 1

moles of MnO2 = 16.0 g / 86.97 g/mol = 0.184 mol

moles of HCl = 30.0 g / 36.46 g/mol = 0.823 mol

from the balanced equation 1 mol of MnO2 required 4 mol fo HCl accordingly

0.184 mol of MnO2 required 4 x 0.184 mol = 0.736 mol

but we have 0.823 mol means we have excess

a.

limiting agent is MNO2

we have to calculate the yield with respect to limiting agent

b.

again from balanced equation

1 mol of MnO2 will give 1 mol of MnCl2 accordingly

0.184 mol of MnO2 will give 0.184 mol of MnCl2

mass of MnCl2 = moles x molar mass = 0.184 mol x 125.844 g/mol = 23.155 g/mol

c.

weh reaction is completed HCl is remaining = 0.823 - 0.736 = 0.087 mol

mass of HCl remaining = 0.087 mol x 36.46 g/mol = 3.172 g

d.

% of yield = [actual yield / theritical yield ] x 100

71.2 % = [ actual yield / 23.155] x 100

Actual yield = 71.2 x 23.155 / 100 = 16.48 g