In general before one attempts to balance the masses in a chemical reaction is necessary to determine first if there is a change of oxidation state (* ) from the reactants to the product of the reaction..If there is a change the total units of oxidation-reduction must be first balanced before balancing the mass.

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is in the -3 oxidation state ( because it is combined with 3H+) this means there is a reduction of 3 units per N atom and a total of 6 for the 2 atoms reacting.. now H is in the 0 oxidation state also and is oxidized to the +1 state in the product NH3 and since there are 2 of them the total oxidation units are2.
A reaction requires the number of oxidation units to be equal to the reduction units. We can obtain this by using 3 molecules of H . If we use 3H2,the equation will be balanced in redox units.
There are 2 N on the left and only one on the right. a coefficient of 2 for NH3 will satisfy the mass requirements ( 2N,left; 2N right; 6H left; 6H right )
(*) the oxidation state of an atom is the number of valence electrons shared or exchanged to produce the chemical bond between to different atoms. Uncombined atoms or atoms of the same elements combined to form a molecule are considered to be in the 0 oxidation state.
Potassium is a metal ( electropositive ) and monovalenns .since Cl is combined with one K it most behave as monovalent negati have , hence Cl is in the -1 oxidation state
one of the negative valence of oxygen is compensated by one positive valence of H and the remaining 7 negative valences must be compensated by Cl. In this case then Cl is in the +7 oxidation state.

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S is bound to 4O ( -8 total negative valences ) and 2 H ( 2 positive valences ) the 6 free negative valences must be compensated by S .In this compound S is in the +6 oxidation state.

Balancing chemical equations is usually easier than it looksThere are 2 Nitrogens and 2 Hydrogens on the left sideThere are 1 Nitrogen and 3 Hydrogens on the right sideHydrogen seems to be the more complicated term so we should start with that_N2 + _H2 ⇒ _NH3
With 2 H on one side and 3 on the other, our common multiple is 6To balance the H we need to multiply H2 by 3 and NH3 by 2_N2 + 3H2 ⇒ 2NH3
Now that we have H balanced, we need to look at NThere are 2 Nitrogens on both sides of the equation, so in this case it is done
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