Step 2: compare the number of atoms that reactants with the number of atoms of products.

You are watching: S+hno3=h2so4+no2+h2o

Reactants (left side)Products (right side)
ElementNumber that atomsNumber that atoms
S11
H14
O37
N11

Step 3: Now, first let us take into consideration nitrogen atom. If we multiply 6 in the reactant (HNO3) and product (in NO2), us will acquire the equal variety of atoms.

No. Of atoms of nitrogenReactant (in HNO3)Product (in NO2)
Initially11
To balance1 x 6 = 61 x 6 = 6

Step 4: write the result equation:

S + 6HNO3→ H2SO4 + 6NO2 + H2O

Step 5: Now check whether the equation is well balanced or no by to compare the atoms

Reactants (left side)Products (right side)
ElementNumber that atomsNumber the atoms
S11
H64
O1817
N66

We uncover that the equation is not well balanced yet. Together the variety of hydrogen and also oxygen atoms are unequal top top the 2 sides.

First balance the oxygen atom.

Step 6: If us multiply 2 in the product (in H2O) and 6, us will acquire the equal variety of oxygen atom on both sides.

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No. Of atoms of nitrogenReactant (in HNO3)Product (in H2SO4+ + 6NO2+ H2O)
Initially1817
To balance184 + 12 + 2 x 1 = 18

Step 7: create the resulting equation:

S + 6HNO3→ H2SO4 + 6NO2 + 2H2O

Step 8: Now inspect whether the equation is balanced or no by comparing the atoms.

Reactants (left side)Products (right side)
ElementNumber the atomsNumber of atoms
S11
H66
O1818
N66

Step 9: create the result equation: