<u>Answer:</u> The mass of the excess reactant (oxygen gas) is 3.136 grams
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
.....(1)
Given mass of HCl = 53.2 g
Molar mass of HCl = 36.5 g/mol
Putting values in equation 1, we get:

Given mass of oxygen gas = 26.5 g
Molar mass of oxygen gas = 32 g/mol
Putting values in equation 1, we get:

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

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 =
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:

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