In the following chemical reaction between H_2 and Cl_2 to produce HCl, what is the mass of HCl produced and leftover reactants

Question

4 years ago

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when 0.36 g of H_2 completely reacts with 12.41 g of Cl_2?I

1 answer:

ira [324] · Answer 1 · 2022-02-12T11:33:57+03:00

Answer: The total amount of leftover reactants and HCl is 12.79 grams

Explanation:

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

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

Given mass of hydrogen gas = 0.36 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen gas}=\frac{0.36g}{2g/mol}=0.18mol$

Given mass of chlorine gas = 12.41 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

$\text{Moles of chlorine gas}=\frac{12.41g}{71g/mol}=0.175mol$

The chemical equation for the reaction of hydrogen gas and chlorine gas is:

$H_2+Cl_2\rightarrow 2HCl$

By Stoichiometry of the reaction:

1 moles of chlorine gas reacts with 1 mole of hydrogen gas

So, 0.175 moles of chlorine gas will react with = $\frac{1}{1}\times 0.175=0.175mol$ of hydrogen gas

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

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

Moles of excess reactant left (hydrogen gas) = [0.18 - 0.175] = 0.005 moles

By Stoichiometry of the reaction

1 moles of chlorine gas produces 2 moles of HCl

So, 0.175 moles of chlorine gas will produce = $\frac{2}{1}\times 0.175=0.350$ moles of HCl

Now, calculating the mass of hydrogen gas left and HCl from equation 1, we get:

Molar mass of hydrogen gas = 2 g/mol

Moles of excess hydrogen gas = 0.005 moles

Putting values in equation 1, we get:

$0.005mol=\frac{\text{Mass of excess hydrogen gas}}{2g/mol}\\\\\text{Mass of excess hydrogen gas}=(0.005mol\times 2g/mol)=0.01g$

Molar mass of HCl = 36.5 g/mol

Moles of HCl = 0.350 moles

Putting values in equation 1, we get:

$0.350mol=\frac{\text{Mass of HCl}}{36.5g/mol}\\\\\text{Mass of HCl}=(0.350mol\times 36.5g/mol)=12.78g$

Total mass of HCl and leftover reactants = [12.78 + 0.01] = 12.79 g

Hence, the total amount of leftover reactants and HCl is 12.79 grams