Respuesta :
Answer: a) The maximum mass of sulfur trioxide that can be formed is 31.4 grams
b) The FORMULA for the limiting reagent is [tex]O_2[/tex]
c) Mass of excess reagent remains is 4.8 grams
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} SO_2=\frac{29.9g}{64g/mol}=0.467moles[/tex]
[tex]\text{Moles of} O_2=\frac{6.26g}{32g/mol}=0.196moles[/tex]
[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)[/tex]
According to stoichiometry :
1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]
Thus 0.196 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.196=0.392moles[/tex] of [tex]SO_2[/tex]
Thus [tex]O_2[/tex] is the limiting reagent as it limits the formation of product and [tex]SO_2[/tex] is the excess reagent as (0.467-0.392) = 0.075 moles or [tex]0.075mol\times 64g/mol=4.8g[/tex] are left.
As 1 mole of [tex]O_2[/tex] give = 2 moles of [tex]SO_3[/tex]
Thus 0.196 moles of [tex]O_2[/tex] give =[tex]\frac{2}{1}\times 0.196=0.392moles[/tex] of [tex]SO_3[/tex]
Mass of [tex]SO_3=moles\times {\text {Molar mass}}=0.392moles\times 80g/mol=31.4g[/tex]
Thus 31.4 g of [tex]SO_3[/tex] will be produced from the given masses of both reactants.
