How would you balance this equation??
Ba(NO3)2 + NH2SO3H + H2O ->
Ba(NH2SO3)2 + HNO3
Would you have to make the H2O 0, ultimately removing H2O from the equation (is that even allowed?), or is there another way of balancing it??
To balance this equation, we need to make sure that the number of atoms on both sides of the equation is the same.
Here's the step-by-step process to balance this equation:
1. Start by balancing the atoms that appear in only one compound on each side of the equation. In this case, we see that the only compound with sulfur (S) is NH2SO3H on the left side, so let's start with that.
Ba(NO3)2 + NH2SO3H + H2O -> Ba(NH2SO3)2 + HNO3
2. Balance the sulfur (S) atoms by placing the coefficient 2 in front of NH2SO3H on the left side:
Ba(NO3)2 + 2NH2SO3H + H2O -> Ba(NH2SO3)2 + HNO3
3. Next, balance the nitrogen (N) atoms. There are two nitrate ions (NO3-) on the left side, so we need two molecules of NH2SO3H on the right side to balance the nitrogen:
Ba(NO3)2 + 2NH2SO3H + H2O -> Ba(NH2SO3)2 + HNO3
4. Now, let's balance the barium (Ba) atoms. There is one barium atom on the left side and two on the right side, so we need a coefficient of 2 in front of Ba(NO3)2:
2Ba(NO3)2 + 2NH2SO3H + H2O -> Ba(NH2SO3)2 + HNO3
5. Finally, balance the hydrogen (H) and oxygen (O) atoms. On the left side, there are 2 hydrogen atoms in the H2O molecule, which means we need a coefficient of 2 in front of HNO3 on the right side:
2Ba(NO3)2 + 2NH2SO3H + H2O -> Ba(NH2SO3)2 + 2HNO3
Now, the equation is balanced with the same number of atoms on both sides. There is no need to remove water (H2O) from the equation because it is a reactant that participates in the reaction.