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Answer:
The above equation is a neutralisation reaction between Sodium Hydroxide and Sulphuric acid (base and acid). As a result of the reaction between an acid and a base, salt (Sodium Sulphate) and water is produced. But the molecules of each element are not the same on both sides. Therefore we will use a step by step method to balance the equation.
- The first step is to create a table and check the number of atoms of each element on each side of the equation. Remember we have to see each element separately.
Name of element | No. of molecules on LHS | No. of molecules on RHS |
Na (Sodium) | 1 | 2 |
O (Oxygen) | 1 ( in NaOH) +4 (in H2SO4) = 5 | 4 (in Na2SO4)+1= 5 |
H (Hydrogen) | 3 (1 in NaOH and 2 in H2SO4) | 2 |
S (Sulphur) | 1 | 1 |
Now we see that Oxygen and sulphur have the same number of molecules on each side. So we start by balancing H, which has the highest number of unequal molecules.
- Since H is three on LHS and two on RHS, we will give one more H on RHS. But since H2 is combined, we have to multiply by 2.
The result is as below: NaOH + H2SO4 → Na2SO4+2H2O
Name of element | No. of molecules on LHS | No. of molecules on RHS |
Na (Sodium) | 1 | 2 |
O (Oxygen) | 1 ( in NaOH) +4 (in H2SO4) = 5 | 6 (because by multiplying H2O with 2 Oxygen became 6 (4+2) |
H (Hydrogen) | 3 (1 in NaOH and 2 in H2SO4) | 4 |
S (Sulphur) | 1 | 1 |
- Now we can see that O has the maximum count of unbalanced atoms.
So we’ll multiply NaOH by two so that 2+4 becomes six and the count equalises.
So the equation becomes 2NaOH + H2SO4 → Na2SO4+2H2O.
Now the table is:
Name of element | No. of molecules on LHS | No. of molecules on RHS |
Na (Sodium) | 2 | 2 |
O (Oxygen) | 2 ( in NaOH) +4 (in H2SO4) = 6 | 4 (in Na2SO4)+2= 6 |
H (Hydrogen) | 4 (2 in NaOH and 2 in H2SO4) | 4 |
S (Sulphur) | 1 | 1 |
- Now, as we can see in the table, all the elements have been balanced. So the right answer is:
2NaOH + H2SO4 → Na2SO4+2H2O
