Question

SO₂ + H₂S → 3S + H₂O The above reaction is balanced or not.​

Answer 1

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } SolutioN:-}}}}}$

Given reaction is of Sulphur dioxide with Hydrogen di sulphide which results in the formation of sulphur and water .

$\boxed{\red{\bf{\leadsto \:\:\underset{\blue{Sulphur\: dioxide}}{SO_2}+\underset{\blue{Hydrogen\:sulphide}}{H_2S}\longrightarrow \underset{\blue{Sulphur}}{3S}+\underset{\blue{Water}}{H_2O}}}}$

Now lets see number of Atoms on LHS and RHS ,

$\begin{tabular}{|c|c|c|} \cline{1-3} \bf Atom & \bf RHS & \bf LHS \\\cline{1-3} \sf Sulphur & 2 & 3 \\\cline{1-3}\sf Hydrogen & 2 & 2 \\\cline{1-3} \sf Oxygen & 2 & 1 \\ \cline{1-3} \end{tabular}$

Hence we can see that the Number of Atoms on LHS and RHS is not equal , So the reaction is not balanced .

Now after balancing ,

$\boxed{\red{\bf{\leadsto \:\:\underset{\blue{Sulphur\: dioxide}}{SO_2}+\underset{\blue{Hydrogen\:sulphide}}{2H_2S}\longrightarrow \underset{\blue{Sulphur}}{3S}+\underset{\blue{Water}}{2H_2O}}}}$

$\begin{tabular}{|c|c|c|} \cline{1-3} \bf Atom & \bf RHS & \bf LHS \\\cline{1-3} \sf Sulphur & 2\times 1+1=3 & 3 \\\cline{1-3}\sf Hydrogen & 2\times2=4 & 2\times2=4 \\\cline{1-3} \sf Oxygen & 2 & 2\times1=2\\ \cline{1-3} \end{tabular}$

Hence there are equal number of atoms on LHS and RHS . So , now we can say that the equation is balanced .

$\boxed{\green{\bf \pink{\dag} \:Hence\:the\:eqution\:is\:now\:balanced\:!}}$

Answer 2

Answer:

LHS≠RHS So, the equation is not balanced

Explanation:

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LHS l RHS

l

S = 1 l S = 3

H = 2 l H = 2

O = 2 l O = 1

I

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So, the given chemical equation is not following the rule of "law of conservation of mass",which states "mass can neither be created nor it can be destroyed "

So, the balanced chemical equation of the above reaction is :

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SO2 + 2H2O------>3S + 2H20

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Hope it helps