Respuesta :
Answer: The formula of limiting reagent is 'S', the amount of excess reagent (carbon monoxide) left is 3.22 grams and the maximum amount of sulfur dioxide formed is 23.2 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 sulfur:
Given mass of sulfur = 11.6 g
Molar mass of sulfur = 32 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of sulfur}=\frac{11.6g}{32g/mol}=0.3625mol[/tex]
- For carbon monoxide:
Given mass of carbon monoxide = 23.5 g
Molar mass of carbon monoxide = 28 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of carbon monoxide}=\frac{23.5g}{28g/mol}=0.840mol[/tex]
The chemical equation for the reaction of sulfur and carbon monoxide follows:
[tex]S(s)+2CO(g)\rightarrow SO_2(g)+2C(s)[/tex]
By Stoichiometry of the reaction:
1 mole of sulfur reacts with 2 moles of carbon monoxide
So, 0.3625 moles of sulfur will react with = [tex]\frac{2}{1}\times 0.3625=0.725mol[/tex] of carbon monoxide
As, given amount of carbon monoxide is more than the required amount. So, it is considered as an excess reagent.
Thus, sulfur is considered as a limiting reagent because it limits the formation of product.
Amount of excess reagent (carbon monoxide) = [0.840 - 0.725] = 0.115 moles
By Stoichiometry of the reaction:
1 mole of sulfur produces 1 mole of sulfur dioxide
So, 0.3625 moles of sulfur will produce = [tex]\frac{1}{1}\times 0.3625=0.3625moles[/tex] of sulfur dioxide
Now, calculating the mass of carbon monoxide and sulfur dioxide from equation 1, we get:
- For carbon monoxide:
Molar mass of carbon monoxide = 28 g/mol
Excess moles of carbon monoxide = 0.115 moles
Putting values in equation 1, we get:
[tex]0.115mol=\frac{\text{Mass of carbon monoxide}}{28g/mol}\\\\\text{Mass of carbon monoxide}=(0.115mol\times 28g/mol)=3.22g[/tex]
- For sulfur dioxide:
Molar mass of sulfur dioxide = 64 g/mol
Excess moles of sulfur dioxide = 0.3625 moles
Putting values in equation 1, we get:
[tex]0.3625mol=\frac{\text{Mass of sulfur dioxide}}{64g/mol}\\\\\text{Mass of sulfur dioxide}=(0.3625mol\times 64g/mol)=23.2g[/tex]
Hence, the formula of limiting reagent is 'S', the amount of excess reagent (carbon monoxide) left is 3.22 grams and the maximum amount of sulfur dioxide formed is 23.2 grams
