Question

How many grams of Cl2 (MW = 71.0 g/mol) can be prepared from the reaction of 16.0 g of MnO2 (FW = 86.9 g/mol) and 30.0 g of HCl (FW = 36.5 g/mol) according to the following chemical equation?MnO2 + 4 HCl ® MnCl2 + Cl2 + 2 H2O0.82 g5.8 g13.0 g14.6 g58.4 g

Answer 1

Answer: The mass of chlorine produced in the given reaction is 13.206 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For manganese oxide:

Given mass of manganese oxide = 16.0 g

Molar mass of manganese oxide = 86.9 g/mol

Putting values in equation 1, we get:

$\text{Moles of manganese oxide}=\frac{16g}{86.9g/mol}=0.184mol$

• For hydrochloric acid:

Given mass of hydrochloric acid = 30.0 g

Molar mass of hydrochloric acid = 36.5 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrochloric acid}=\frac{30g}{36.5g/mol}=0.822mol$

For the given chemical reaction:

$MnO_2+4HCl\rightarrow MnCl_2+Cl_2+2H_2O$

By Stoichiometry of the reaction:

1 mole of manganese oxide reacts with 3 moles of hydrochloric acid.

So, 0.184 moles of manganese oxide will react with = $\frac{3}{1}\times 0.184=0.736moles$ of hydrochloric acid

As, given amount of hydrochloric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, manganese oxide is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of manganese oxide reacts with 1 moles of chlorine.

So, 0.184 moles of manganese oxide will react with = $\frac{1}{1}\times 0.184=0.184moles$ of chlorine.

Now, calculating the mass of chlorine from equation 1, we get:

Molar mass of chlorine = 71 g/mol

Moles of chlorine = 0.186 moles

Putting values in equation 1, we get:

$0.186mol=\frac{\text{Mass of chlorine}}{71g/mol}\\\\\text{Mass of chlorine}=13.206$

Hence, the mass of chlorine produced in the given reaction is 13.206 grams.