Respuesta :
Answer: There will be 131.3 tons of [tex] SO_{2} [/tex] that will be generated.
Explanation : We have the data as 72 tons of [tex]H_{2}S[/tex], [tex]O_{2}[/tex] is 101 tons and [tex]H_{2}O[/tex] as 38 tons.
Conversion of 1 ton to kg will be approximately 907.2 Kg
We have to convert the given data of tons into kg for all;
So, moles of [tex]H_{2}S[/tex] = (72 tons X 907.2 Kg/ton X 1000 g/Kg) / (34.082 g/ mol) = 1.916 X [tex]10^{6}[/tex] moles
Now, moles of [tex]O_{2}[/tex] = (101 tons X 907.2 Kg/ton X 1000 g/Kg) / (32 g/mol) = 2.80 X[tex]10^{6}[/tex] moles
And, [tex]H_{2}O[/tex] = (38 tons X 907.2 Kg/ton X 1000 g/Kg) / (18.016 g/mol)
= 1.03 X [tex]10^{6}[/tex] moles
Now, we have to calculate the moles of [tex]H_{2}S[/tex]
so, 1.916 X [tex]10^{6}[/tex] moles / 2 moles of [tex]H_{2}S[/tex] = 958 X [tex]10^{3}[/tex]
Now, for [tex]O_{2}[/tex] ;
So, 2.80 X[tex]10^{6}[/tex] moles / 3 moles of [tex]O_{2}[/tex] = 933.3 X [tex]10^{3}[/tex] moles
Here, we can see that [tex]O_{2}[/tex] is acting as a limiting reactant.
Now, moles of [tex]SO_{2}[/tex] = 2.80 X[tex]10^{6}[/tex] X (2 moles of [tex]SO_{2}[/tex] / 3 moles of [tex]O_{2}[/tex]) = 1.86 X [tex]10^{6}[/tex]
Now converting this into mass,
Mass = Moles X Molar Mass
Mass = 1.86 X [tex]10^{6}[/tex] X (64.066 g/mol X [tex]10^{-3}[/tex] kg/g) X (1 ton / 907.2 Kg)
= 131.3 tons of [tex]SO_{2}[/tex]
Answer:
135 tons of [tex]SO_2[/tex] are formed
Explanation:
[tex]2H_2S+3O_2\rightarrow 2H_2O+2SO_2[/tex]
Every 72 tons of [tex]H_2S[/tex] reacts with 101 tons of [tex]O_2[/tex] and forms 38 tons of [tex]H_2O[/tex] and x amount of [tex]SO_2[/tex].
As per as law of conservation of mass:
Total mass of reactants = Total mass of products
72 tons + 101 tons = 38 tons + x
x = 135 tons
135 tons of [tex]SO_2[/tex] are formed
