Question

What is the total mass of products formed when 34 g of H2S is oxidised by excess of oxygen gas to
produce water & sulphur dioxide?​

Answer 1

Answer : The total mass of products formed is, 82 grams

Explanation : Given,

Mass of $H_2S$ = 34 g

Molar mass of $H_2S$ = 34 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $SO_2$ = 64 g/mole

First we have to calculate the moles of $H_2S$

$\text{Moles of }H_2S=\frac{\text{Mass of }H_2S}{\text{Molar mass of }H_2S}=\frac{34g}{34g/mole}=1mole$

Now we have to calculate the moles of $H_2O$ and $SO_2$

The balanced chemical reaction will be,

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

From the balanced chemical reaction we conclude that,

As, 2 mole of $H_2S$ react to give 2 mole of $H_2O$

So, 1 mole of $H_2S$ react to give 1 mole of $H_2O$

and

As, 2 mole of $H_2S$ react to give 2 mole of $SO_2$

So, 1 mole of $H_2S$ react to give 1 mole of $SO_2$

Now we have to calculate the mass of $H_2O$ and $SO_2$

$\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O=1mole\times 18g/mole=18g$

$\text{Mass of }SO_2=\text{Moles of }SO_2\times \text{Molar mass of }SO_2=1mole\times 64g/mole=64g$

Now we have to calculate the total mass of products.

Total mass of products = Mass of $H_2O$ + Mass of $SO_2$

Total mass of products = 18 g + 64 g = 82 g

Therefore, the total mass of products are, 82 grams

Answer 2

Answer: The total mass of the products obtained is 82 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$       ......(1)

Given mass of hydrogen sulfide = 34 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in above equation, we get:

$\text{Moles of }H_2S=\frac{34g}{34g/mol}=1mol$

The chemical equation for the reaction of hydrogen sulfide and oxygen follows:

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

As, oxygen is present in excess, so it is considered as an excess reagent.

So, hydrogen sulfide is considered as a limiting reagent because it limits the formation of products.

• For water:

By Stoichiometry of the reaction:

2 moles of hydrogen sulfide produces 2 moles of water.

So, 1 mole of hydrogen sulfide will produce = $\frac{2}{2}\times 1=1mol$

Calculating the mass of water from equation 1, we get:

Moles of water = 1 mole

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

$1mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=18g$

Mass of water produced = 18 grams

• For sulfur dioxide:

By Stoichiometry of the reaction:

2 moles of hydrogen sulfide produces 2 moles of sulfur dioxide.

So, 1 mole of hydrogen sulfide will produce = $\frac{2}{2}\times 1=1mol$ of sulfur dioxide.

Calculating the mass of sulfur dioxide from equation 1, we get:

Moles of sulfur dioxide = 1 mole

Molar mass of sulfur dioxide = 64 g/mol

Putting values in equation 1, we get:

$1mol=\frac{\text{Mass of sulfur dioxide}}{64g/mol}\\\\\text{Mass of sulfur dioxide}=64g$

Mass of sulfur dioxide produced = 64 grams

• Total mass of the products = 18 g + 64 g = 82 g

Hence, the total mass of the products obtained is 82 grams.