Question

Hydrogen chloride gas and oxygen gas react to form water and chlorine gas. A reaction mixture initially contains 53.2 g of hydrogen chloride and 26.5 g of oxygen gas. Once the reaction has occurred as completely as possible, what mass (in g) of the excess reactant remains? Enter to 1 decimal place.

Answer 1

Answer: The mass of the excess reactant (oxygen gas) is 3.136 grams

Step-by-step explanation:

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

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

• For HCl:

Given mass of HCl = 53.2 g

Molar mass of HCl = 36.5 g/mol

Putting values in equation 1, we get:

`\text{Moles of HCl}=(53.2g)/(36.5g/mol)=1.46mol`

• For oxygen gas:

Given mass of oxygen gas = 26.5 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

`\text{Moles of oxygen gas}=(26.5g)/(32g/mol)=0.828mol`

The chemical equation for the reaction of HCl and oxygen gas follows:

`2HCl+O_2\rightarrow H_2O+Cl_2`

By Stoichiometry of the reaction:

2 moles of HCl reacts with 1 mole of oxygen gas

So, 1.46 moles of HCl will react with =
`(1)/(2)* 1.46=0.73mol` of oxygen gas

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

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

Excess moles of oxygen gas = (0.828 - 0.73) = 0.098 moles

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

Molar mass of oxygen gas = 32 g/mol

Excess moles of oxygen gas = 0.098 moles

Putting values in equation 1, we get:

`0.098mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.098mol* 32g/mol)=3.136g`

Hence, the mass of the excess reactant (oxygen gas) is 3.136 grams