How do you balance C3H5(NO3)3 -->CO2 + H20 + N2 + O2?

Answer 1

#2C_3H_5(NO_3)_3# #rarr# #6CO_2# + #5H_2O# + #3N_2# + #1/2 O_2#

Because of the large molecule on your left hand side and the potential effects of placing a coefficient ahead of it on all those atoms, this is a little tricky to balance.

Totaling all the atoms involved is the first step.

#C_3H_5(NO_3)_3# #rarr# #CO_2# + #H_2O# + #N_2# + #O_2# (unbalanced)

According to the subscipts, we possess

left side

#C# = 3 #H# = 5 #N# = 3 #O# = 9
Notice that since the #NO_3# was enclosed by a parenthesis followed by the subscript 3, I have to multiply the number of atoms inside the parenthesis by 3. (e.g. #N_"1 x 3"# ; #O_"3 x 3"#)

right side

#C# = 1 #H# = 2 #N# = 2 #O# = 2 + 1 + 2 (do not add this up yet)

I'm going to choose a component from the right side that I believe will be the key to balancing the entire equation because the molecules there are far simpler than the molecules on the left.

#C_3H_5(NO_3)_3# #rarr# #CO_2# + #color (blue) 5H_2O# + #N_2# + #O_2#

left side

#C# = 3 #H# = 5 #N# = 3 #O# = 9

right side

#C# = 1 #H# = 2 x #color (blue) 5# = 10 #N# = 2 #O# = 2 + (1 x #color (blue) 5#) + 2
Notice that since #H_2O# is a substance, I would also need to multiply the #O# by 5. Now that there are 10 atoms of #H# on the right side, I would need to also have 10 atoms of #H# on the left side.
#color (red) 2C_3H_5(NO_3)_3# #rarr# #CO_2# + #5H_2O# + #N_2# + #O_2#

left side

#C# = 3 x #color (red) 2# = 6 #H# = 5 x #color (red) 2# = 10 #N# = 3 x #color (red) 2# = 6 #O# = 9 x #color (red) 2# = 18

right side

#C# = 1 #H# = 2 x 5 = 10 #N# = 2 #O# = 2 + (1 x 5) + 2

Now let's balance the remainder of the equation, and once more note that all of the atoms on the left side are multiplied by 2 because they are bonded to one another.

#2C_3H_5(NO_3)_3# #rarr# #color (green) 6CO_2# + #5H_2O# + #color (orange) 3N_2# + #O_2#

left side

Numbers C, H, and N are equivalent to 3 x 2 = 6, 5 x 2 = 10, and 9 x 2 = 18.

right side

#C# = 1 x #color (green) 6# = 6 #H# = 2 x 5 = 10 #N# = 2 x #color (orange) 3# = 6 #O# = (2 x #color (green) 6#) + (1 x 5) + 2
Now the only atom left to balance is the #O#. Notice that if you add all the O atoms on the right side, you will come up with a sum of 19. So what to do? Use your knowledge of fractions.
#2C_3H_5(NO_3)_3# #rarr# #6CO_2# + # 5H_2O# + #3N_2# + #color (magenta) (1/2)O_2#

left side

Numbers C, H, and N are equivalent to 3 x 2 = 6, 5 x 2 = 10, and 9 x 2 = 18.

right side

#C# = 1 x 6 = 6 #H# = 2 x 5= 10 #N# = 2 x 3 = 6 #O# = (2 x 6) + (1 x 5) + (2 x #color (magenta) (1/2)#) = 18

At this point, the equation is balanced.

But you can always multiply the WHOLE equation by two if you prefer whole number coefficients to fractions.

#cancel 2# [#2C_3H_5(NO_3)_3# #rarr# #6CO_2# + #5H_2O# + #3N_2# + #1/ cancel 2O_2#]

=

#4C_3H_5(NO_3)_3# #rarr# #12CO_2# + #10H_2O# + #6N_2# + #O_2#
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Answer 2

The balanced equation for the given reaction is:

[ C_3H_5(NO_3)_3 \rightarrow 3CO_2 + 2H_2O + 3N_2 + 5O_2 ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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