Respuesta :
Answer :
(a) The limiting reactant is, [tex]H_2O[/tex]
(b) The mass of [tex]C_2H_6O_2[/tex] is, 11570 grams
(c) The mass of [tex]C_2H_6O_2[/tex] is, 5785 grams
Explanation : Given,
Molar rate of [tex]C_2H_4O[/tex] = 80 kg/s = 80000 g/s
Molar rate of [tex]H_2O[/tex] = 25.5 lbm/s = 11.57 kg/s = 11570 g/s
conversion used : (1 lbm = 0.453592 kg)
Molar mass of [tex]C_2H_4O[/tex] = 44 g/mole
Molar mass of [tex]H_2O[/tex] = 18 g/mole
Molar mass of [tex]C_2H_6O_2[/tex] = 50 g/mole
First we have to calculate the moles of [tex]C_2H_4O[/tex] and [tex]H_2O[/tex] per second.
[tex]\text{Moles of }C_2H_4O=\frac{\text{Mass of }C_2H_4O}{\text{Molar mass of }C_2H_4O}=\frac{80000g}{44g/mole}=1818.18mole/s[/tex]
[tex]\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{11570g}{50g/mole}=231.4mole/s[/tex]
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
[tex]C_2H_4O(g)+H_2O(g)\rightarrow C_2H_6O_2(g)[/tex]
From the balanced reaction we conclude that
The mole ratio of [tex]C_2H_4O[/tex] and [tex]H_2O[/tex] is, 1 : 1
From this we conclude that, [tex]C_2H_4O[/tex] is an excess reagent because the given moles are greater than the required moles and [tex]H_2O[/tex] is a limiting reagent and it limits the formation of product.
Now we have to calculate the moles of [tex]C_2H_6O_2[/tex].
As, 1 moles of [tex]H_2O[/tex] react to give 6 moles of [tex]C_2H_6O_2[/tex]
So, 231.4 moles of [tex]H_2O[/tex] react to give 231.4 moles of [tex]C_2H_6O_2[/tex]
Now we have to calculate the mass of [tex]C_2H_6O_2[/tex].
[tex]\text{Mass of }C_2H_6O_2=\text{Moles of }C_2H_6O_2\times \text{Molar mass of }H_2O[/tex]
[tex]\text{Mass of }C_2H_6O_2=(231.4mole)\times (50g/mole)=11570g[/tex]
Now we have to calculate the mass of of [tex]C_2H_6O_2[/tex].
As we are given that [tex]C_2H_4O[/tex] is 50 % pure.
So, the mass of [tex]C_2H_4O[/tex] = [tex]11570g\times \frac{50}{100}=5785g[/tex]
The mass of [tex]C_2H_6O_2[/tex] is, 5785 grams
