Question

The poisonous gas hydrogen sulfide, H2S, can be neutralized with a base such as sodium hydroxide,

NaOH. The unbalanced equation for this reaction follows:
NaOH(aq) + H25(e) → Na2S(aq) + H2O(1)
A student who was asked to balance this equation wrote the following:
Na2OH(aq) + H2S(g) → Na2S(aq) + H2O(
Is this equation balanced? Is it correct? Explain why or why not, and supply the correct balanced
equation if necessary.

Answer 1

The student's equation is incorrect because the chemical formula for sodium hydroxide is NaOH, not Na2OH, and the reaction is not balanced. The correct balanced equation for the neutralization of hydrogen sulfide with sodium hydroxide is 2 NaOH(aq) + H2S(g) → Na2S(aq) + 2 H2O(l).

Step-by-step explanation:

The equation provided by the student is not correct and not balanced. The correct formula for sodium hydroxide is NaOH, not Na2OH, which does not exist as a compound. Also, in the reaction of hydrogen sulfide (H2S) with sodium hydroxide (NaOH), two moles of sodium hydroxide react with one mole of hydrogen sulfide to produce sodium sulfide (Na2S) and water (H2O). Therefore, the balanced chemical equation is:

2 NaOH(aq) + H2S(g) → Na2S(aq) + 2 H2O(l)

Here, the law of conservation of mass is satisfied, as we have two sodium (Na) atoms, two hydroxide (OH) groups, one sulfur (S) atom, and two hydrogen (H) atoms on both sides of the equation.

Answer 2

It's not correct. For balancing, we need to put the coefficients in the molecule, not in the athom. Because if you do this, you're creating another molecule, instead of a balacing, for which the reaction may not happen - but anyway, it would be another reaction.

The correct balacing is:

2NaOH + 1H2S → 1Na2S + 2H2O

Step-by-step explanation:

Look: Na2OH does not even exist. OH has only one free link, so he can't - in normal conditions - make another one with any athom. That's why we should write 2NaOH instead of Na2OH. The first means "2 mols of NaOH".