Question

balance e following chemical equation SO2 +H2S = S+H2O

Answer 1

Answer:

SO2+ 2HS gives 2H2O+ 3S

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Answer 2

Answer:

The balanced equation for the given chemical reaction is

$\boxed{\sf\:SO_{2\:(\:g\:)}\:+\:2\:H_2S_{aq}\:\longrightarrow\:3\:S_{\:(\:s\:)}\:+\:2\:H_2O_{(\:l\:)}}$

Explanation:

To balance the given equation, we have to follow the following steps.

Step 1:

Rewrite the given equation as it is.

$\sf\:SO_{2\:(\:g\:)}\:+\:H_2S_{(\:aq\:)}\:\longrightarrow\:S_{(\:g\:)}\:+\:H_2O_{(\:l\:)}$

Step 2:

Write the number of atoms of each of the elements in the given equation on the both sides of the given equation.

$\begin{array}{|c|c|c|}\cline{1-3}\sf\:Element & \sf\:No.\:of\:atoms\:in\:reactant & \sf\:No.\:of\:atoms\:in\:product\\\cline{1-3}\sf\:S & \sf\:2 & \sf\:1\\\cline{1-3}\sf\:O & \sf\:2 & \sf\:1\\\cline{1-3}\sf\:H & \sf\:2 & \sf\:2\\\cline{1-3}\end{array}$

From the above table, we know that,

The number of hydrogen atoms on both sides is same.

We have to now balance the number of sulphur and oxygen atoms.

Step 3:

Balance the number of sulphur atoms:

$\begin{array}{|c|c|c|}\cline{1-3}\sf\:No.\:of\:sulphur\:atoms & \sf\:In\:reactants & \sf\:In\:products\:\\\cline{2-3}& \sf\:SO_2\:\:\:\:H_2S & \sf\:(\:S\:)\\\cline{1-3}\sf\:Initially & \sf\:1\:\:\:\:1 & \sf\:1\\\cline{1-3}\sf\:To\:balance & \sf\:1\:\:\:\:1 & \sf\:1\:\times\:2\\\cline{1-3}\end{array}$

To balance the number of sulphur atoms in the product, we have to multiply it by 2.

Now, the equation becomes

$\sf\:SO_2\:+\:H_2S\:\longrightarrow\:2\:S\:+\:H_2O$

Step 4:

Balance the number of oxygen atoms:

$\begin{array}{|c|c|c|}\cline{1-3}\sf\:No.\:of\:oxygen\:atoms & \sf\:In\:reactants\:(\:SO_2\:) & \sf\:In\:products\:(\:H_2O\:)\\\cline{1-3}\sf\:Initially & \sf\:2 & \sf\:1\\\cline{1-3}\sf\:To\:balance & \sf\:2 & \sf\:1\:\times\:2\\\cline{1-3}\end{array}$

To balance the number of oxygen atoms in product, we have to multiply it by 2.

Now, the equation becomes-

$\sf\:SO_2\:+\:H_2S\:\longrightarrow\:2\:S\:+\:2\:H_2O$

Step 5:

Balance the number of hydrogen atoms in the unbalanced equation:

$\begin{array}{|c|c|c|}\cline{1-3}\sf\:No.\:of\:hydrogen\:atoms & \sf\:In\:reactants\:(\:H_2S\:) & \sf\:In\:products\:(\:H_2O\:)\\\cline{1-3}\sf\:Initially & \sf\:2 & \sf\:4\\\cline{1-3}\sf\:To\:balance & \sf\:2\:\times\:2 & \sf\:4\\\cline{1-3}\end{array}$

To balance the number of hydrogen atoms in reactant, we have to multiply it by 2.

Now, the equation becomes-

$\sf\:SO_2\:+\:2\:H_2S\:\longrightarrow\:2\:S\:+\:2\:H_2O$

From this equation, we know that, the number of sulphur atoms in the product side are one less than in the reactant side.

To balance it, we have to add one number to the number of sulphur atoms in product side i. e. 2S + 1S = 3S

Hence, now the equation becomes -

$\sf\:SO_2\:+\:2\:H_2S\:\longrightarrow\:3\:S\:+\:2\:H_2O$

Now, this equation is balanced.

To make it more informative, we can mention the physical states of the elements.

The balanced equation is -

$\boxed{\sf\:SO_{2\:(\:g\:)}\:+\:2\:H_2S_{aq}\:\longrightarrow\:3\:S_{\:(\:s\:)}\:+\:2\:H_2O_{(\:l\:)}}$