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velikii [3]
3 years ago
9

Hydrogen chloride gas and oxygen gas react to form water and chlorine gas. A reaction mixture initially contains 53.2 g of hydro

gen 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.
Chemistry
1 answer:
Sergio [31]3 years ago
4 0

<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:

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

  • <u>For HCl:</u>

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}=\frac{53.2g}{36.5g/mol}=1.46mol

  • <u>For oxygen gas:</u>

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}=\frac{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 = \frac{1}{2}\times 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\times 32g/mol)=3.136g

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

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