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)

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)

the awnser is group 12