Answer to Question #126570 in Chemistry for rita

Question

Hydrogen sulphide reacts with oxygen to produce sulphur dioxide and water. Determine the mass of sulphur dioxide formed if 5.00 g of hydrogen sulphide reacts with excess oxygen. Clearly show your method of calculation.

Answer 1

The equation of the reaction between the hydrogen sulfide and oxygen is:

2H₂S+ 3O₂ $\leftrightarrow$ 2SO₂ + 2H₂O.

According to the stoichiometric coefficients , when 2 molecules of hydrogen sulfide react, two molecules of sulfur oxide are formed. Therefore, the number of the moles of the hydrogen sulfide reacted equals the number of the moles of the sulfur oxide formed:

$n(H_2S) =n(SO_2)$ .

The number of the moles of hydrogen sulfide contained in 5 g can be calculated making use the number of the moles and molar mass $M$ relation (34.1 g/mol for hydrogen sulfide).

$n=m/M$ .

$n=5/34.1 = 0.147$ mol.

Therefore, the number of the moles of the sulfur dioxide produced is 0.147 mol. The molar mass of sulfur dioxide is 64.067 g/mol. Finally, the mass of the sulfur dioxide is:

$m=nM = 0.147•64.067= 9.4$ g.

Answer: 9.4 g of sulfur dioxide is formed when 5 g of hydrogen sulfide react with oxygen.