Balance the equation (nh4)3 po4 +na0h arrow na3p04 +3nh3 +3h20.

Explanation:

(NH4)3 PO4 +NaOH arrow Na3PO4 +3NH3 +3H2O

Start by seeing what happens with the Na. You need 3 on the left, so put a 3 in front of NaOH

(NH4)3 PO4 +3NaOH arrow Na3PO4 +3NH3 +3H2O  Next work with the nitrogens. YOu have 3 on the left and 3 on the right, so they are OK. Next Go to the stray oxygens.

You have 3 on left in (NaOH) and three on the right in 3H2O so they are fine as well. The last thing you should look at are hydrogens.

There are 12 + 3 on the left which is 15. There are 9 (in 3NH3) and 6 more in the water. They seem fine.

Why didn't I do something with the PO4^(-3)? The reason is a deliberately stayed away from them and balanced everything else. Since they were untouched with 1 on the left and 1 on the right, they are balanced.

Species      Na        H        O         N       PO4

Left             3          15        3         3          1

Right           3         15         3         3          1

(NH₄)₃PO₄ + 3NaOH ⟶ Na₃PO₄ + 3NH₃ + 3H₂O

Explanation:

Your unbalanced equation is

(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + NH₃ + H₂O

A method that usually works for balancing by inspection is

1. Pick the most complicated-looking formula [(NH₄)₃PO₄].

Put a 1 in front of it.

1(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + NH₃ + H₂O

2. Balance N.

We have fixed 3N on the left. We need 3N on the right.

Put a 3 in front of NH₃.

1(NH₄)₃PO₄ + NaOH ⟶ Na₃PO₄ + 3NH₃ + H₂O

3. Balance P.

We have fixed 1P on the left. We need 1P on the right.

Put a 1 in front of Na₃PO₄.

1(NH₄)₃PO₄ + NaOH ⟶ 1Na₃PO₄ + 3NH₃ + H₂O

4. Balance Na

We have fixed 3Na on the right. we need 3Na on the left.

Put a 3 in front of NaOH.

1(NH₄)₃PO₄ + 3NaOH ⟶ 1Na₃PO₄ + 3NH₃ + H₂O

5. Balance O.

We have fixed 7O on the left and 4O on the right. We need three more O atoms on the right.

Put a 3 in front of H₂O.

1(NH₄)₃PO₄ + 3NaOH ⟶ 1Na₃PO₄ + 3NH₃ + 3H₂O

All species have a coefficient. The equation should now be balanced.

6. Check that all atoms are balanced

The balanced equation is

(NH₄)₃PO₄ + 3NaOH ⟶ Na₃PO₄ + 3NH₃ + 3H₂O

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.