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Niacin (#C_6NH_5O_2#) reacts with #O_2# to produce #CO_2#, #H_2O#, and #NO_2#. How would you write and balance the equation for this reaction?

Answer 1

#4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2#

After adjusting the no of atoms of 6C,1N and 5 H in RHS with LHS ,we get total #(14+5/2) # O at RHS. To adjust the no of O atom in LHS we can write #(6+5/4) #before #O_2# in LHS and the balanced equation becomes.

#C_6NH_5O_2+(6+5/4)O_2->6CO_2+5/2H_2O+NO_2#

After removing the fraction by multiplying both sides by 4, we are left with the following balanced equation.

#4C_6NH_5O_2+29O_2->24CO_2+10H_2O+4NO_2#

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Answer 2

Noah Aldrich

6 C6H5NO2 + 47 O2 → 12 CO2 + 10 H2O + 2 NO2

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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.