Balancing C3H5N3>O9 ⇒ CO2 + N2 + H2O + O2

➤

Chemical Reactions ➤
Balancing C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

Balance C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + __O₂

Our reactant has all uneven numbers, whereas our products' are all even
let's fix that by multiplying the reactant by 2

2 C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

and now we adjust the products accordingly
leaving O till last

2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + __ O₂

element count

balancing chem

6 C
10 H
6 N
18 O

balancing chemical equation

6 C
10 H
6 N
19 O

chem equations

obviously, the O is giving us grief...
we have an extra O on the product side

So, how do we fix this? By cutting the O₂ in half
1/2 O₂ = 1 O

2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + 1/2 O₂

now, we can't go to the lab and meassure out 1/2 mole of O₂
that's why we want to keep the coefficients as whole numbers at all times

the easiest way to fix this mess is to multiply the whole equation by 2

2 (2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + 1/2 O₂)

=

4 C₃H₅N₃O₉ ⇒ 12 CO₂ + 6 N₂ + 10 H₂O + 1 O₂

element count

balancing chem

12 C
20 H
12 N
36 O

balancing chemical equation

12 C
20 H
12 N
36 O

Academic Subjects

Chemistry

General Chemistry

Chemical Reactions ➤
Balancing C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

Balancing C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

Balance C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + __O₂

Our reactant has all uneven numbers, whereas our products' are all even
let's fix that by multiplying the reactant by 2

2 C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

and now we adjust the products accordingly
leaving O till last

2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + __ O₂

element count

balancing chem

6 C
10 H
6 N
18 O

balancing chemical equation

6 C
10 H
6 N
19 O

chem equations

balancing chem

6 C
10 H
6 N
18 O

balancing chemical equation

6 C
10 H
6 N
19 O

chem equations

obviously, the O is giving us grief...
we have an extra O on the product side

So, how do we fix this? By cutting the O₂ in half
1/2 O₂ = 1 O

2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + 1/2 O₂

now, we can't go to the lab and meassure out 1/2 mole of O₂
that's why we want to keep the coefficients as whole numbers at all times

the easiest way to fix this mess is to multiply the whole equation by 2

2 (2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + 1/2 O₂)

=

4 C₃H₅N₃O₉ ⇒ 12 CO₂ + 6 N₂ + 10 H₂O + 1 O₂

element count

balancing chem

12 C
20 H
12 N
36 O

balancing chemical equation

12 C
20 H
12 N
36 O

balancing chem

12 C
20 H
12 N
36 O

balancing chemical equation

12 C
20 H
12 N
36 O

Academic Subjects

Chemistry

General Chemistry

Chemical Reactions ➤ Balancing C3H5N3O9 ⇒ CO2 + N2 + H2O + O2

Balancing C3H5N3O9 ⇒ CO2 + N2 + H2O + O2

Balance C3H5N3O9 ⇒ CO2 + N2 + H2O + O2

__C3H5N3O9 ⇒ __CO2 + __N2 + __H2O + __O2

Our reactant has all uneven numbers, whereas our products' are all even let's fix that by multiplying the reactant by 2

2 C3H5N3O9 ⇒ __CO2 + __N2 + __H2O + __O2

and now we adjust the products accordinglyleaving O till last

2 C3H5N3O9 ⇒ 6 CO2 + 3 N2 + 5 H2O + __ O2

element count balancing chem 6 C 10 H 6 N 18 O balancing chemical equation 6 C 10 H 6 N 19 O chem equations

balancing chem

6 C 10 H 6 N 18 O

balancing chemical equation

6 C 10 H 6 N 19 O

chem equations

obviously, the O is giving us grief... we have an extra O on the product side So, how do we fix this? By cutting the O2 in half 1/2 O2 = 1 O

2 C3H5N3O9 ⇒ 6 CO2 + 3 N2 + 5 H2O + 1/2 O2

now, we can't go to the lab and meassure out 1/2 mole of O2 that's why we want to keep the coefficients as whole numbers at all times the easiest way to fix this mess is to multiply the whole equation by 2

2 (2 C3H5N3O9 ⇒ 6 CO2 + 3 N2 + 5 H2O + 1/2 O2) = 4 C3H5N3O9 ⇒ 12 CO2 + 6 N2 + 10 H2O + 1 O2

element count balancing chem 12 C 20 H 12 N 36 O balancing chemical equation 12 C 20 H 12 N 36 O $(function(){ $("#fot-placeholder").load("footer.html"); });

balancing chem

12 C 20 H 12 N 36 O

balancing chemical equation

12 C 20 H 12 N 36 O

Chemical Reactions ➤
Balancing C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

Balancing C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

Balance C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + __O₂

Our reactant has all uneven numbers, whereas our products' are all even
let's fix that by multiplying the reactant by 2

2 C₃H₅N₃O₉ ⇒ CO₂ + N₂ + H₂O + O₂

and now we adjust the products accordingly
leaving O till last

2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + __ O₂

element count

balancing chem

6 C
10 H
6 N
18 O

balancing chemical equation

6 C
10 H
6 N
19 O

chem equations

6 C
10 H
6 N
18 O

6 C
10 H
6 N
19 O

obviously, the O is giving us grief...
we have an extra O on the product side

So, how do we fix this? By cutting the O₂ in half
1/2 O₂ = 1 O

2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + 1/2 O₂

now, we can't go to the lab and meassure out 1/2 mole of O₂
that's why we want to keep the coefficients as whole numbers at all times

the easiest way to fix this mess is to multiply the whole equation by 2

2 (2 C₃H₅N₃O₉ ⇒ 6 CO₂ + 3 N₂ + 5 H₂O + 1/2 O₂)

=

4 C₃H₅N₃O₉ ⇒ 12 CO₂ + 6 N₂ + 10 H₂O + 1 O₂

element count

balancing chem

12 C
20 H
12 N
36 O

balancing chemical equation

12 C
20 H
12 N
36 O

12 C
20 H
12 N
36 O

12 C
20 H
12 N
36 O