Question

How many grams of C12 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according to the following chemical equation? Mn02 + 4HCI → MnCl2 + Cl2 + 2H2O

Answer 1

Mass of Cl₂ produced 12.78 g

Step-by-step explanation:

Given data:

Mass of MnO₂ = 16 g

Mass of HCl = 30.0 g

Mass of Cl₂ produced = ?

Solution:

Chemical equation:

MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

Number of moles of MnO₂:

Number of moles = mass / molar mass

Number of moles = 16 g/ 87 g/mol

Number of moles = 0.18 mol

Number of moles of HCl:

Number of moles = mass / molar mass

Number of moles = 30 g/ 36.46 g/mol

Number of moles = 0.82 mol

Now we will compare the moles of Cl₂ with MnO₂ and HCl.

MnO₂ : Cl₂

1 : 1

0.18 : 0.18

HCl : Cl₂

4 : 1

0.82 : 1/4×0.82 = 0.205 mol

The number of moles of Cl₂ formed by HCl are less it will limiting reactant.

Mass of Cl₂ formed:

Mass = number of moles × molar mass

Mass = 0.18 mol × 71 g/mol

Mass = 12.78 g