How do you balance the equation for this reaction: #C(s) +SO_2(g) -> CS_2(l) + CO_2(g)#?

2 Answers
May 10, 2017

Answer:

#3C(s) + 2SO_2 (g) → CS_2 (l) + 2CO_2 (g)#

Explanation:

#C(s) + SO_2 (g) → CS_2 (l) + CO_2 (g)#

We have:
LHS
C= 1
S= 1
O=2

RHS
C=2
S=2
O=2

We balance the equation by making the number of each element the same on both sides.

So:
LHS
C= #1xx3# = 3
S= #1xx2#= 2
O= #2xx2#= 4

RHS
C= 1 (from #CS_2#) + #(2xx1)# (from the #CO_2#) = 3
S= 2
O=#2xx2=4#

Then we rewrite the equation:

#3C(s) + 2SO_2 (g) → CS_2 (l) + 2CO_2 (g)#

Hope this isn't too confusing!

May 10, 2017

Answer:

#3C(s) + 2SO_2(g) -> CS_2(l) + 2CO_2(g)#

Explanation:

Given: #C(s) + SO_2(g) -> CS_2(l) + CO_2(g)#

According to the Law of Conservation of mass, the number of atoms of each type on the left side of the equation must be equal to the number of atoms on the right side of the equation:

#ul(Left" " Side) " " ul(Right" " Side)#
#C - 1 " " C - 2#
#S - 1 " " S - 2#
#O - 2" " O - 2#

You can't change formulas, only add a different number of molecules (coefficient). Since the Carbon atoms are in two different molecules on the right, start with balancing the Sulfur:

#C(s) + 2SO_2(g) -> CS_2(l) + CO_2(g)#

#ul(Left" " Side) " " ul(Right" " Side)#
#C - 1 " " C - 2#
#S - 2 " " S - 2#
#O - 4" " O - 2#

Add a #2# in front of the #CO_2# to balance the Oxygen:

#C(s) + 2SO_2(g) -> CS_2(l) + 2CO_2(g)#

#ul(Left" " Side) " " ul(Right" " Side)#
#C - 1 " " C - 3#
#S - 2 " " S - 2#
#O - 4" " O - 4#

Finally, put a #3# in front of the single Carbon:

#3C(s) + 2SO_2(g) -> CS_2(l) + 2CO_2(g)#

#ul(Left" " Side) " " ul(Right" " Side)#
#C - 3 " " C - 3#
#S - 2 " " S - 2#
#O - 4" " O - 4#

Balance!