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QUESTION

# Translate this equation-hydrogen sulphide gas burns in air to give water and sulphur dioxide and then balance them?

2H_2S_((g)) + 3O_(2(g)) -> 2SO_(2(g)) + 2H_2O_((g))

The unbalanced combustion reaction of hydrogen sulfide, H_2S, looks like this

H_2S_((g)) + O_(2(g)) -> SO_(2(g)) + H_2O_((g))

This is a highly , so the water will be produced as steam. To balance this equation, you need to match the number of atoms present on the reactants' side with the number of atoms present on the products' side.

You have 1 sulfur atom on the reactants' side and 1 on the products' side, so sulfur is balanced.

Notice that you have 2 oxygen atoms on the reactants' side, and 3 on products' side. In order to balance the oxygen atoms, multiply the oxygen molecule by 3, the sulfur dioxide molecule by 2, and the water molecule by 2

H_2S_((g)) + color(red)(3)O_(2(g)) -> color(blue)(2)SO_(2(g)) + color(green)(2)H_2O_((g))

The oxygen atoms are now balanced - you have 6 on the lefthand side of the equation and 6 on the righthand side of the equation.

Notice that the sulfur atoms are no longer balanced. To balance the sulfur again, multiply the hydrogen sulfide molecule by 2

2H_2S_((g)) + color(red)(3)O_(2(g)) -> color(blue)(2)SO_(2(g)) + color(green)(2)H_2O_((g))

The hydrogen atoms are balanced, since you ahve 4 on the lefthand side and *48 on the righthand side of the equation.

All the atoms are balanced, so this is how the balanced chemical equation looks like

2H_2S_((g)) + color(red)(3)O_(2(g)) -> color(blue)(2)SO_(2(g)) + color(green)(2)H_2O_((g))