Question

.  If 1 g of HCl and 1 g of MnO2 heated together the maximum weight of Cl2 gas evolved will be :
[MnO2 + 4HCl ⊗ MnCl2 + Cl2 + 2H2O] :​

Answer 1

Answer: 0.35 g

Explanation: To calculate the amount of water produced, we first need to find the limiting reagent and for it we need to find the number of moles. To calculate the moles,we use the following equation:  

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$  

Moles of HCl:

Putting values in above equation, we get:

$\text{Number of moles of HCl}=\frac{1g}{36.5g/mol}=0.03moles$

Moles of $MnO_2$

Putting values in above equation, we get:

$\text{Number of moles}=\frac{1g}{87g/mol}=0.01moles$

For the given chemical reaction, the equation follows:

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

By Stoichiometry:

4 moles of  hydrochloric acid reacts with 1 mole of manganese oxide

So, 0.03 moles of hydrochloric acid reacts with = $\frac{1}{4}\times 0.03=0.0075moles$ of managnese oxide

As, the required amount of  manganese oxide is less than the given amount. Hence, it is considered as the excess reagent.

Hydrochloric acid is considered as a limiting reagent because it limits the formation of product.

4 moles of hydrochloric acid produces 1 mole of chlorine

Thus 0.03 moles of hydrochloric acid produces =$\frac{1}{4}\times 0.03=0.0075$ moles of chlorine

Mass of chlorine=$moles\times {\text {Molar mass}}=0.0075\times 71=0.53g$