Question

1. _h₂s + _cl₂ → _s₈ + _hcl
2. _fe + _h₂o → _fe₃o₄ + _h₂
3. _n₂ + _h₂ → ___nh₃
4. _n₂ + _o₂ → ___n₂o
5. _co₂ + _h₂o → _c₆h₁₂o₆ + _o₂
6. _sicl₄ + _h₂o → _h₄sio₄ + _hcl
7. _h₃po₄ → _h₄p₂o₇ + ___h₂o

Explanation:

Step1: Balance the sulfur atoms

For the equation \( \text{H}_2\text{S}+\text{Cl}_2 ightarrow\text{S}_8+\text{HCl}\), there are 8 sulfur atoms in \( \text{S}_8\). So we put 8 in front of \( \text{H}_2\text{S}\), getting \(8\text{H}_2\text{S}+\text{Cl}_2 ightarrow\text{S}_8+\text{HCl}\).

Step2: Balance the hydrogen atoms

Since there are \(8\times2 = 16\) hydrogen atoms from \(8\text{H}_2\text{S}\), we put 16 in front of \( \text{HCl}\), resulting in \(8\text{H}_2\text{S}+\text{Cl}_2 ightarrow\text{S}_8 + 16\text{HCl}\).

Step3: Balance the chlorine atoms

There are 16 chlorine atoms in \(16\text{HCl}\), so we put 8 in front of \( \text{Cl}_2\). The balanced equation is \(8\text{H}_2\text{S}+8\text{Cl}_2=\text{S}_8 + 16\text{HCl}\).

Step4: For \( \text{Fe}+\text{H}_2\text{O}

ightarrow\text{Fe}_3\text{O}_4+\text{H}_2\)
Balance the iron atoms: put 3 in front of \( \text{Fe}\), getting \(3\text{Fe}+\text{H}_2\text{O} ightarrow\text{Fe}_3\text{O}_4+\text{H}_2\).

Step5: Balance the oxygen atoms

There are 4 oxygen atoms in \( \text{Fe}_3\text{O}_4\), so put 4 in front of \( \text{H}_2\text{O}\), resulting in \(3\text{Fe}+4\text{H}_2\text{O} ightarrow\text{Fe}_3\text{O}_4+\text{H}_2\).

Step6: Balance the hydrogen atoms

There are \(4\times2 = 8\) hydrogen atoms from \(4\text{H}_2\text{O}\), so put 4 in front of \( \text{H}_2\). The balanced equation is \(3\text{Fe}+4\text{H}_2\text{O}=\text{Fe}_3\text{O}_4 + 4\text{H}_2\).

Step7: For \( \text{N}_2+\text{H}_2

ightarrow\text{NH}_3\)
Balance the nitrogen atoms: there are 2 nitrogen atoms in \( \text{N}_2\), so put 2 in front of \( \text{NH}_3\), getting \( \text{N}_2+\text{H}_2 ightarrow2\text{NH}_3\).

Step8: Balance the hydrogen atoms

There are \(2\times3 = 6\) hydrogen atoms in \(2\text{NH}_3\), so put 3 in front of \( \text{H}_2\). The balanced equation is \( \text{N}_2+3\text{H}_2 = 2\text{NH}_3\).

Step9: For \( \text{N}_2+\text{O}_2

ightarrow\text{N}_2\text{O}\)
Balance the oxygen atoms: there is 1 oxygen atom in \( \text{N}_2\text{O}\), so put 2 in front of \( \text{N}_2\text{O}\), getting \( \text{N}_2+\text{O}_2 ightarrow2\text{N}_2\text{O}\).

Step10: Balance the nitrogen atoms

There are \(2\times2 = 4\) nitrogen atoms in \(2\text{N}_2\text{O}\), so put 2 in front of \( \text{N}_2\). Now balance the oxygen atoms: there are 2 oxygen atoms in \( \text{O}_2\) and 2 oxygen atoms in \(2\text{N}_2\text{O}\). The balanced equation is \(2\text{N}_2+\text{O}_2 = 2\text{N}_2\text{O}\).

Step11: For \( \text{CO}_2+\text{H}_2\text{O}

ightarrow\text{C}_6\text{H}_{12}\text{O}_6+\text{O}_2\)
Balance the carbon atoms: put 6 in front of \( \text{CO}_2\), getting \(6\text{CO}_2+\text{H}_2\text{O} ightarrow\text{C}_6\text{H}_{12}\text{O}_6+\text{O}_2\).

Step12: Balance the hydrogen atoms

There are 12 hydrogen atoms in \( \text{C}_6\text{H}_{12}\text{O}_6\), so put 6 in front of \( \text{H}_2\text{O}\), resulting in \(6\text{CO}_2+6\text{H}_2\text{O} ightarrow\text{C}_6\text{H}_{12}\text{O}_6+\text{O}_2\).

Step13: Balance the oxygen atoms

There are \(6\times2+6\times1 = 18\) oxygen atoms on the left - hand side and \(6 + 2x\) (where \(x\) is the coefficient of \( \text{O}_2\)) on the right - hand side. Solving \(6 + 2x=18\) gives \(x = 6\). The balanced equation is \(6\text{CO}_2+6\text{H}_2\text{O}=\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\).

Step14: For \( \text{SiCl}_4+\text{H}_2\text{O}

ightarrow\text{H}_4\text{SiO}_4+\text{HCl}\)
Balance the silicon atoms: already balanced.

Step15: Balance the chlorine atoms

There are 4 chlorine atoms in \( \text{SiCl}_4\), so put 4 in front of \( \text{HCl}\), getting \( \text{SiCl}_4+\text{H}_2\text{O} ightarrow\text{H}_4\text{SiO}_4+4\text{HCl}\).

Step16: Balance the hydrogen and oxygen atoms

There are 4 hydrogen atoms in \( \text{H}_4\text{SiO}_4\) and \(4\times1 = 4\) hydrogen atoms in \(4\text{HCl}\), so there are \(4 + 4=8\) hydrogen atoms. Put 4 in front of \( \text{H}_2\text{O}\). The balanced equation is \( \text{SiCl}_4+4\text{H}_2\text{O}=\text{H}_4\text{SiO}_4 + 4\text{HCl}\).

Step17: For \( \text{H}_3\text{PO}_4

ightarrow\text{H}_4\text{P}_2\text{O}_7+\text{H}_2\text{O}\)
Balance the phosphorus atoms: put 2 in front of \( \text{H}_3\text{PO}_4\), getting \(2\text{H}_3\text{PO}_4 ightarrow\text{H}_4\text{P}_2\text{O}_7+\text{H}_2\text{O}\).

Step18: Balance the hydrogen and oxygen atoms

There are \(2\times3 = 6\) hydrogen atoms in \(2\text{H}_3\text{PO}_4\), and \(4 + 2x\) (where \(x\) is the coefficient of \( \text{H}_2\text{O}\)) on the right - hand side. Also, for oxygen: \(2\times4=8\) on the left and \(7 + x\) on the right. Solving \(4 + 2x=6\) (hydrogen balance) gives \(x = 1\), and checking oxygen: \(7+1 = 8\). The balanced equation is \(2\text{H}_3\text{PO}_4=\text{H}_4\text{P}_2\text{O}_7+\text{H}_2\text{O}\).

Answer:

1. \(8\text{H}_2\text{S}+8\text{Cl}_2=\text{S}_8 + 16\text{HCl}\)
2. \(3\text{Fe}+4\text{H}_2\text{O}=\text{Fe}_3\text{O}_4 + 4\text{H}_2\)
3. \( \text{N}_2+3\text{H}_2 = 2\text{NH}_3\)
4. \(2\text{N}_2+\text{O}_2 = 2\text{N}_2\text{O}\)
5. \(6\text{CO}_2+6\text{H}_2\text{O}=\text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2\)
6. \( \text{SiCl}_4+4\text{H}_2\text{O}=\text{H}_4\text{SiO}_4 + 4\text{HCl}\)
7. \(2\text{H}_3\text{PO}_4=\text{H}_4\text{P}_2\text{O}_7+\text{H}_2\text{O}\)