# Solve the following reactions: Au2 S3+H2 -> Hg(OH)2+H3 PO4 -> SiO2+ HF-> B2Br6 + HNO3-> KOH + H3PO4-> (NH4)3PO4 + Pb(NO3)4 ->

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The following reactions are balancing reactions, so:

Au2 S3+H2 -> Au + H2S

In order to balance the equations, we have to have 2 atoms of Au, both side of the equation, so for this reason we have to multiply Au with 2, the result being:

Au2 S3+H2 -> 2Au + H2S

We'll verify if the number of atoms, both side of the equation, is equal and we'll notice that we have 1 atom of S on the right side, instead 3 atoms of S, entered in the equation, so, we'll multiply on the right side, 3*H2S, the equation becoming:

Au2 S3+H2 -> 2Au + 3*H2S

The last step is to balance the number of H entered into equation, for this reason multiplying 3*H2, the final result being:

**Au2 S3+3*H2 -> 2Au + 3*H2S**

Knowing the rule, now we can balance the following equation.

Remember!

The number of atoms entered into equation has to be the same at the exit from the equation!

2) Hg(OH)2+H3 PO4 -> Hg3(PO4)2 + H2O

Now, all we have to do is to balance the equation:

**3Hg(OH)2 + 2H3 PO4 -> Hg3(PO4)2 + 6H2O**

3) SiO2+ HF-> SiF4 + H2O

Balancing the equation, we'll have:

**SiO2 + 4HF-> SiF4 + 2H2O**

4) B2Br6 + HNO3->B(NO3)3 + HBr

We'll do the balance of the equation:

** B2Br6 + 6HNO3-> 2B(NO3)3 + 6HBr**

5) KOH + H3PO4-> K3(PO4) + H2O

Balancing the equation, we'll have:

** 3KOH + H3PO4-> K3(PO4) + 3H2O**

6) (NH4)3PO4 + Pb(NO3)4 -> Pb3(PO4)4+ NH4NO3

Balancing the equation, we'll have:

** 4(NH4)3PO4 + 3Pb(NO3)4 -> Pb3(PO4)4+ 12NH4NO3**

- Au2S3+H2 -> 2Au + 3H2S
- 3Hg(OH)2 + 2H3PO4 -> Hg3(PO4)2 + 6H2O
- SiO2 + 4HF -> SiF4 + 2H2O
- B2 Br6 + 6HNO3 -> 2B(NO3)3 + 6HBr
- 3KOH + H3PO4 -> K3PO4 + 3H2O
- 4(NH4)3 PO4 + 3Pb(NO3)4 -> Pb3(PO4)4+ 12NH4NO3