Respuesta :
Answer: The total amount of leftover reactants and HCl is 12.79 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
- For hydrogen gas:
Given mass of hydrogen gas = 0.36 g
Molar mass of hydrogen gas = 2 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of hydrogen gas}=\frac{0.36g}{2g/mol}=0.18mol[/tex]
- For chlorine gas:
Given mass of chlorine gas = 12.41 g
Molar mass of chlorine gas = 71 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of chlorine gas}=\frac{12.41g}{71g/mol}=0.175mol[/tex]
The chemical equation for the reaction of hydrogen gas and chlorine gas is:
[tex]H_2+Cl_2\rightarrow 2HCl[/tex]
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 = [tex]\frac{1}{1}\times 0.175=0.175mol[/tex] 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 = [tex]\frac{2}{1}\times 0.175=0.350[/tex] moles of HCl
Now, calculating the mass of hydrogen gas left and HCl from equation 1, we get:
- For hydrogen gas:
Molar mass of hydrogen gas = 2 g/mol
Moles of excess hydrogen gas = 0.005 moles
Putting values in equation 1, we get:
[tex]0.005mol=\frac{\text{Mass of excess hydrogen gas}}{2g/mol}\\\\\text{Mass of excess hydrogen gas}=(0.005mol\times 2g/mol)=0.01g[/tex]
- For HCl:
Molar mass of HCl = 36.5 g/mol
Moles of HCl = 0.350 moles
Putting values in equation 1, we get:
[tex]0.350mol=\frac{\text{Mass of HCl}}{36.5g/mol}\\\\\text{Mass of HCl}=(0.350mol\times 36.5g/mol)=12.78g[/tex]
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
