Insert the missing coefficients in the following partially balanced chemical equations s+6hno=hso4+no+h2o

S + 6HNO₃ ⟶ H₂SO₄ +  6NO₂ + 2H₂O

Step-by-step explanation:

S + 6HNO₃ ⟶ H₂SO₄ +  NO₂ + H₂O  

One way is to balance this equation is the oxidation number method.

Step 1. Calculate the oxidation number of every atom:

S⁰ + H⁺¹N⁺⁵O₃⁻² ⟶ H₂⁺¹S⁺⁶O₄⁻² + N⁺⁴O₂⁻² + H₂⁺¹O⁻²

Step 2. Identify the changes in oxidation number:

S: +0 ⟶ +6; Change = +6

N: +5 ⟶ +4; Change =   -1

Step 3. Equalize the changes in oxidation number.

You need 6 atoms of N for every 1 atom of S. This gives us total changes of -6 and + 6.

Step 4. Insert coefficients to get these numbers.

1S + 6HNO₃ ⟶ 1H₂SO₄ +  6NO₂ + H₂O

Step 5. Balance O.

We have fixed 18 O on the left and 16 O on the right. We need 2 more O on the right. Put a 2 in front of H₂O.

1S + 6HNO₃ ⟶ 1H₂SO₄ +  6NO₂ + 2H₂O

Every formula now has a coefficient. The equation should be balanced.

Step 6. Check that all atoms balance.

Atom  Left-hand side  Right-hand side  

 S                  1                           1

 H                 6                          6

 N                 6                          6

 O               18                          18

The balanced equation is

S + 6HNO₃ ⟶ H₂SO₄ +  6NO₂ + 2H₂O

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2