# For the reaction , $\text{C}+\text{xHN}{{\text{O}}_{3}}\left( \text{conc}\text{.} \right)\to \text{C}{{\text{O}}_{2}}+\text{4N}{{\text{O}}_{2}}+\text{y}{{\text{H}}_{2}}\text{O}$ , the value of x and y will be:

A. x= 4, y=2

B. x= 1, y=2

C. x= 2, y=4

D. x= 4, y=4

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**Hint:**Count the number of atoms of every element in the reactants side and the number of atoms of every element in the products side. If the number of atoms are same then, it is balanced otherwise change the coefficients of the molecules until the number of atoms of every element on both the sides of the equation are balanced.

**Complete step by step answer:**

Now, let us balance this equation given in the question, $\text{C}+\text{xHN}{{\text{O}}_{3}}\left( \text{conc}\text{.} \right)\to \text{C}{{\text{O}}_{2}}+\text{4N}{{\text{O}}_{2}}+\text{y}{{\text{H}}_{2}}\text{O}$. We will balance the equation step by step,

Step 1- Examine the number of atoms of different elements in the reaction.

Element | Number of atoms in reactants side (LHS) | Number of atoms in products side (RHS) |

C | 1 | 1 |

H | x | 2y |

N | x | 4 |

O | 3x | 10+y |

Step 2- Try to balance atoms of elements other than oxygen and hydrogen, like N and C. We can see that carbon is already balanced. The element nitrogen can be easily as the atoms of nitrogen in LHS is x and in RHS is 4. So, to equalize the two, x should be equal to 4. So, x=4. To make four nitrogen atoms in LHS, we need to multiply $\text{HN}{{\text{O}}_{3}}$ with 4. The number of atoms after multiplying $\text{HN}{{\text{O}}_{3}}$ with 4 will be

Element | Number of atoms in reactants side (LHS) | Number of atoms in products side (RHS) |

C | 1 | 1 |

H | 4 | 2y |

N | 4 | 4 |

O | 12 | 10+y |

Step 3- Carbon and nitrogen are already balanced; now just find the number of atoms of H and O need to be balanced. To balance water, we need to find the value of y. Then, water will be balanced. In hydrogen, 2y=4, so, y=2. Then, the number of atoms of hydrogen and oxygen are -

Element | Number of atoms in reactants side (LHS) | Number of atoms in products side (RHS) |

C | 1 | 1 |

H | 4 | 4 |

N | 4 | 4 |

O | 12 | 10+2=12 |

The equation is balanced now. The balanced equation is by putting x=4 and y=2 is $\text{C}+4\text{HN}{{\text{O}}_{3}}\left( \text{conc}\text{.} \right)\to \text{C}{{\text{O}}_{2}}+\text{4N}{{\text{O}}_{2}}+2{{\text{H}}_{2}}\text{O}$.

**The correct answer of this question is option ‘a’ which is x=4 and y=2.**

**Note:**

The use of balancing an equation is that it helps us to find the amount of reactants needed and the amount of products formed. Every equation must follow the law of conservation of mass. It also states that matter cannot be created nor it can be destroyed, it can only be transformed in one form to another.