Question

Some scientists believe that life on Earth may have originated near deep-ocean vents. Balance the following reactions, which are among those taking place near such vents. Place coefficients directly in front of the compound, leaving no spaces.
CH3SH(l) + CO (g) → CH3CO(SCH3)(l) +H2S (g)
H2S(g) +CO(g) → CH3CO2H(g)+ S8(s)

Answer 1

Answer :

(1) The balanced chemical reaction will be,

$2CH_3SH(l)+CO(g)\rightarrow CH_3CO(SCH_3)(l)+H_2S(g)$

(2) The balanced chemical reaction will be,

$8H_2S(g)+8CO(g)\rightarrow 4CH_3CO_2H(g)+S_8(s)$

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

If the amount of atoms of each type on the left and right sides of a reaction differs then to balance the equation by adding coefficient in the front of the elements or molecule or compound in the chemical equation.

The coefficient tell us about that how many molecules or atoms present in the chemical equation.

(1) The unbalanced chemical reaction is,

$CH_3SH(l)+CO(g)\rightarrow CH_3CO(SCH_3)(l)+H_2S(g)$

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen, sulfur and carbon atoms are not balanced.

In order to balance the chemical equation, the coefficient '2' put before the $CH_3SH$ and we get the balanced chemical equation.

The balanced chemical reaction will be,

$2CH_3SH(l)+CO(g)\rightarrow CH_3CO(SCH_3)(l)+H_2S(g)$

(2) The unbalanced chemical reaction is,

$H_2S(g)+CO(g)\rightarrow CH_3CO_2H(g)+S_8(s)$

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen, sulfur and carbon atoms are not balanced.

In order to balance the chemical equation, the coefficient '8' put before the $H_2S\text{ and }CO$ and the coefficient '4' put before the $CH_3CO_2H$ we get the balanced chemical equation.

The balanced chemical reaction will be,

$8H_2S(g)+8CO(g)\rightarrow 4CH_3CO_2H(g)+S_8(s)$

Answer 2

Answer:

2CH3SH(l) + CO (g) → CH3CO(SCH3)(l) +H2S (g)

8H2S(g) + 8CO(g) → 4CH3CO2H(g)+ S8(s)

Explanation:

Step 1: The unbalanced equation

CH3SH(l) + CO (g) → CH3CO(SCH3)(l) +H2S (g)

Step 2: Balancing the equation

CH3SH(l) + CO (g) → CH3CO(SCH3)(l) +H2S (g)

On the left side we have 1x S, on the right side we have 2x S (1x in CH3CO(SCH3) and 1x in H2S). To balance the amount of S, we have to multiply CH3SH on the left side by 2. Now the equation is balanced.

2CH3SH(l) + CO (g) → CH3CO(SCH3)(l) +H2S (g)

Step 1: The unbalanced equation

H2S(g) +CO(g) → CH3CO2H(g)+ S8(s)

Step 2: Balancing the equation

H2S(g) +CO(g) → CH3CO2H(g)+ S8(s)

On the left side we have 1x S and on the right side we have 8x S

To balance the amount of S on both sides we have to multply H2S on the left by 8.

8H2S(g) +CO(g) → CH3CO2H(g)+ S8(s)

On the left side we have 16x H and on the right side we have 4x H

To balance the amount of H on both sides we have to multply CH3CO2H on the right by 4.

8H2S(g) +CO(g) → 4CH3CO2H(g)+ S8(s)

On the left side we have 1x C and on the right side we have 8x C

To balance the amount of C on both sides we have to multply C0 on the left by 8. Now the equation is balanced.

8H2S(g) + 8CO(g) → 4CH3CO2H(g)+ S8(s)