Question

Volcanoes emit much hydrogen sulfide gas, h2s, which reacts with the oxygen in the air to form water and sulfur dioxide, so2. every 72 tons of h2s reacts with 101 tons of oxygen and forms 38 tons of water. how many tons of so2 are formed?

Answer 1

Answer:

135 tons of $SO_2$ are formed

Explanation:

$2H_2S+3O_2\rightarrow 2H_2O+2SO_2$

Every 72 tons of $H_2S$ reacts with 101 tons of $O_2$ and forms 38 tons of $H_2O$ and x amount of $SO_2$.

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 $SO_2$ are formed

Answer 2

Answer: There will be 131.3 tons of $SO_{2}$ that will be generated.

Explanation : We have the data as 72 tons of $H_{2}S$, $O_{2}$ is 101 tons and $H_{2}O$ 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 $H_{2}S$ = (72 tons X 907.2 Kg/ton X 1000 g/Kg) / (34.082 g/ mol) = 1.916 X $10^{6}$ moles

Now, moles of $O_{2}$ = (101 tons X 907.2 Kg/ton X 1000 g/Kg) / (32 g/mol) = 2.80 X$10^{6}$ moles

And, $H_{2}O$ = (38 tons X 907.2 Kg/ton X 1000 g/Kg) / (18.016 g/mol)

= 1.03 X $10^{6}$ moles

Now, we have to calculate the moles of $H_{2}S$

so, 1.916 X $10^{6}$ moles / 2 moles of $H_{2}S$ = 958 X $10^{3}$

Now, for $O_{2}$ ;

So, 2.80 X$10^{6}$ moles / 3 moles of $O_{2}$ = 933.3 X $10^{3}$ moles

Here, we can see that $O_{2}$ is acting as a limiting reactant.

Now, moles of $SO_{2}$ = 2.80 X$10^{6}$ X (2 moles of $SO_{2}$ / 3 moles of $O_{2}$) = 1.86 X $10^{6}$

Now converting this into mass,

Mass = Moles X Molar Mass

Mass = 1.86 X $10^{6}$ X (64.066 g/mol X $10^{-3}$ kg/g) X (1 ton / 907.2 Kg)

= 131.3 tons of $SO_{2}$