**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) | |

Element | Number that atoms | Number that atoms |

S | 1 | 1 |

H | 1 | 4 |

O | 3 | 7 |

N | 1 | 1 |

**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 nitrogen | Reactant (in HNO3) | Product (in NO2) |

Initially | 1 | 1 |

To balance | 1 x 6 = 6 | 1 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) | |

Element | Number that atoms | Number the atoms |

S | 1 | 1 |

H | 6 | 4 |

O | 18 | 17 |

N | 6 | 6 |

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 nitrogen | Reactant (in HNO3) | Product (in H2SO4+ + 6NO2+ H2O) |

Initially | 18 | 17 |

To balance | 18 | 4 + 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) | |

Element | Number the atoms | Number of atoms |

S | 1 | 1 |

H | 6 | 6 |

O | 18 | 18 |

N | 6 | 6 |

**Step 9:** create the result equation: