Question

identifying reactions
use table 1 to identify the reactions below. then, balance each equation.

1. fe + o₂ → fe₂o₃
2. fes + hcl → fecl₂ + h₂s
3. cl₂ + nabr → nacl + br₂
4. c₁₀h₈ + o₂ → co₂ + h₂o + energy
5. agno₃ + hcl → agcl + hno₃
6. alcl₃ → al + cl₂
7. al + h₂so₄ → al₂(so₄)₃ + h₂
8. kclo₃ → kcl + o₂
9. c₃h₈ + o₂ → co₂ + h₂o + energy
10. h₂(g) + o₂(g) → h₂o(g) + energy

Explanation:

Step1: Identify reaction type for \(Fe + O_2

ightarrow Fe_2O_3\)
It is a synthesis reaction (\(A + B ightarrow AB\)).
To balance:
- For Fe: There are 2 Fe atoms in \(Fe_2O_3\). So put 4 in front of Fe (since \(2\times2 = 4\)).
- For O: There are 3 O atoms in \(Fe_2O_3\). So put 3 in front of \(O_2\) (\(3\times2=6\) O atoms) and 2 in front of \(Fe_2O_3\) (\(2\times3 = 6\) O atoms).
The balanced equation is \(4Fe+3O_2 = 2Fe_2O_3\).

Step2: Identify reaction type for \(FeS + HCl

ightarrow FeCl_2+H_2S\)
It is a double - displacement reaction (\(AB + CD ightarrow AD+CB\)).
To balance:
- For Cl: There are 2 Cl atoms in \(FeCl_2\). So put 2 in front of HCl.
The balanced equation is \(FeS + 2HCl=FeCl_2 + H_2S\).

Step3: Identify reaction type for \(Cl_2+NaBr

ightarrow NaCl + Br_2\)
It is a single - displacement reaction (\(A + BC ightarrow AC + B\)).
To balance:
- For Cl: There are 2 Cl atoms in \(Cl_2\). So put 2 in front of NaCl.
- For Na: There are 2 Na atoms in 2NaCl. So put 2 in front of NaBr.
The balanced equation is \(Cl_2+2NaBr = 2NaCl+Br_2\).

Step4: Identify reaction type for \(C_{10}H_8+O_2

ightarrow CO_2+H_2O+\text{energy}\)
It is a combustion reaction (\(C_xH_y+O_2 ightarrow CO_2 + H_2O+\text{energy}\)).
To balance:
- For C: There are 10 C atoms in \(C_{10}H_8\). So put 10 in front of \(CO_2\).
- For H: There are 8 H atoms in \(C_{10}H_8\). So put 4 in front of \(H_2O\).
- For O: There are \(10\times2+4\times1 = 24\) O atoms on the right - hand side. So put 12 in front of \(O_2\).
The balanced equation is \(C_{10}H_8+12O_2 = 10CO_2+4H_2O\).

Step5: Identify reaction type for \(AgNO_3+HCl

ightarrow AgCl+HNO_3\)
It is a double - displacement reaction (\(AB + CD ightarrow AD+CB\)).
The equation is already balanced as the number of each type of atom is the same on both sides.

Step6: Identify reaction type for \(AlCl_3

ightarrow Al+Cl_2\)
It is a decomposition reaction (\(AB ightarrow A + B\)).
To balance:
- For Cl: There are 3 Cl atoms in \(AlCl_3\). So put 2 in front of \(AlCl_3\) (6 Cl atoms) and 3 in front of \(Cl_2\) (6 Cl atoms).
- For Al: There are 2 Al atoms in 2\(AlCl_3\). So put 2 in front of Al.
The balanced equation is \(2AlCl_3=2Al + 3Cl_2\).

Step7: Identify reaction type for \(Al+H_2SO_4

ightarrow Al_2(SO_4)_3+H_2\)
It is a single - displacement reaction (\(A + BC ightarrow AC + B\)).
To balance:
- For Al: There are 2 Al atoms in \(Al_2(SO_4)_3\). So put 2 in front of Al.
- For \(SO_4\): There are 3 \(SO_4\) groups in \(Al_2(SO_4)_3\). So put 3 in front of \(H_2SO_4\).
- For H: There are \(3\times2 = 6\) H atoms in 3\(H_2SO_4\). So put 3 in front of \(H_2\).
The balanced equation is \(2Al+3H_2SO_4=Al_2(SO_4)_3 + 3H_2\).

Step8: Identify reaction type for \(KClO_3

ightarrow KCl+O_2\)
It is a decomposition reaction (\(AB ightarrow A + B\)).
To balance:
- For O: There are 3 O atoms in \(KClO_3\). So put 2 in front of \(KClO_3\) (6 O atoms) and 3 in front of \(O_2\) (6 O atoms).
- For K and Cl: There are 2 K and 2 Cl atoms in 2\(KClO_3\). So put 2 in front of KCl.
The balanced equation is \(2KClO_3=2KCl + 3O_2\).

Step9: Identify reaction type for \(C_3H_8+O_2

ightarrow CO_2+H_2O+\text{energy}\)
It is a combustion reaction (\(C_xH_y+O_2 ightarrow CO_2 + H_2O+\text{energy}\)).
To balance:
- For C: There are 3 C atoms in \(C_3H_8\). So put 3 in front of \(CO_2\).
- For H: There are 8 H atoms in \(C_3H_8\). So put 4 in front of \(H_2O\).
- For O: There are \(3\times2+4\times1=10\) O atoms on the right - hand side. So put 5 in front of \(O_2\).
The balanced equation is \(C_3H_8+5O_2 = 3CO_2+4H_2O\).

Step10: Identify reaction type for \(H_2(g)+O_2(g)

ightarrow H_2O(g)+\text{energy}\)
It is a synthesis reaction (\(A + B ightarrow AB\)).
To balance:
- For O: There are 2 O atoms in \(O_2\). So put 2 in front of \(H_2O\).
- For H: There are 4 H atoms in 2\(H_2O\). So put 2 in front of \(H_2\).
The balanced equation is \(2H_2+O_2 = 2H_2O\).

Answer:

1. Synthesis, \(4Fe + 3O_2=2Fe_2O_3\)
2. Double - displacement, \(FeS+2HCl = FeCl_2+H_2S\)
3. Single - displacement, \(Cl_2+2NaBr=2NaCl + Br_2\)
4. Combustion, \(C_{10}H_8+12O_2=10CO_2 + 4H_2O\)
5. Double - displacement, \(AgNO_3+HCl=AgCl+HNO_3\)
6. Decomposition, \(2AlCl_3=2Al+3Cl_2\)
7. Single - displacement, \(2Al + 3H_2SO_4=Al_2(SO_4)_3+3H_2\)
8. Decomposition, \(2KClO_3=2KCl+3O_2\)
9. Combustion, \(C_3H_8+5O_2=3CO_2+4H_2O\)
10. Synthesis, \(2H_2+O_2=2H_2O\)