Let's balance each of the chemical equations step-by-step:
### 1. Balancing [tex]\( CH_4 + O_2 \rightarrow CO_2 + H_2O \)[/tex]:
- Step 1: Write down the number of each type of atom present in the unbalanced equation.
- Reactants: [tex]\( C = 1 \)[/tex], [tex]\( H = 4 \)[/tex], [tex]\( O = 2 \)[/tex]
- Products: [tex]\( C = 1 \)[/tex], [tex]\( H = 2 \)[/tex], [tex]\( O = 3 \)[/tex]
- Step 2: Balance the carbon atoms.
- Carbon atoms are already balanced (1 carbon atom on each side).
- Step 3: Balance the hydrogen atoms.
- To balance 4 hydrogen atoms in [tex]\( CH_4 \)[/tex], we need 2 water molecules, [tex]\( 2H_2O \)[/tex].
[tex]\[
CH_4 + O_2 \rightarrow CO_2 + 2H_2O
\][/tex]
- Now, hydrogen: [tex]\( H = 4 \)[/tex] on both sides.
- Step 4: Balance the oxygen atoms.
- On the product side, we have [tex]\( 2 \)[/tex] oxygen atoms in [tex]\( CO_2 \)[/tex] and [tex]\( 2 \times 1 = 2 \)[/tex] in [tex]\( 2H_2O \)[/tex], totaling [tex]\( 4 \)[/tex] oxygen atoms.
- Thus, we need [tex]\( 2 \)[/tex] molecules of [tex]\( O_2 \)[/tex] to provide [tex]\( 4 \)[/tex] oxygen atoms.
[tex]\[
CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O
\][/tex]
Hence, the balanced equation is:
[tex]\[
CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O
\][/tex]
### 2. Balancing [tex]\( CaCl_2 + AgNO_3 \rightarrow Ca(NO_3)_2 + AgCl \)[/tex]:
- Step 1: Write down the number of each type of atom present in the unbalanced equation.
- Reactants: [tex]\( Ca = 1 \)[/tex], [tex]\( Cl = 2 \)[/tex], [tex]\( Ag = 1 \)[/tex], [tex]\( N = 1 \)[/tex], [tex]\( O = 3 \)[/tex]
- Products: [tex]\( Ca = 1 \)[/tex], [tex]\( Cl = 1 \)[/tex], [tex]\( Ag = 1 \)[/tex], [tex]\( N = 2 \)[/tex], [tex]\( O = 6 \)[/tex]
- Step 2: Balance the nitrate groups ([tex]\( NO_3 \)[/tex]).
- We have 2 nitrate groups ([tex]\( NO_3 \)[/tex]) in [tex]\( Ca(NO_3)_2 \)[/tex] on the products side, hence we need 2 molecules of [tex]\( AgNO_3 \)[/tex] on the reactants side.
[tex]\[
CaCl_2 + 2AgNO_3 \rightarrow Ca(NO_3)_2 + AgCl
\][/tex]
- Step 3: Balance the silver and chlorine atoms.
- Now, we have [tex]\( 2AgNO_3 \)[/tex], which provides [tex]\( 2Ag \)[/tex] atoms, thus we need [tex]\( 2 \)[/tex] molecules of [tex]\( AgCl \)[/tex] to balance the silver and chlorine atoms.
[tex]\[
CaCl_2 + 2AgNO_3 \rightarrow Ca(NO_3)_2 + 2AgCl
\][/tex]
Hence, the balanced equation is:
[tex]\[
CaCl_2 + 2AgNO_3 \rightarrow Ca(NO_3)_2 + 2AgCl
\][/tex]
### 3. Balancing [tex]\( C_2H_6O + O_2 \rightarrow CO_2 + H_2O \)[/tex]:
- Step 1: Write down the number of each type of atom present in the unbalanced equation.
- Reactants: [tex]\( C = 2 \)[/tex], [tex]\( H = 6 \)[/tex], [tex]\( O = 1 \)[/tex]
- Products: [tex]\( C = 1 \)[/tex], [tex]\( H = 2 \)[/tex], [tex]\( O = 3 \)[/tex]
- Step 2: Balance the carbon atoms.
- There are 2 carbon atoms in [tex]\( C_2H_6O \)[/tex], so we need 2 molecules of [tex]\( CO_2 \)[/tex] on the products side.
[tex]\[
C_2H_6O + O_2 \rightarrow 2CO_2 + H_2O
\][/tex]
- Step 3: Balance the hydrogen atoms.
- We have 6 hydrogen atoms in [tex]\( C_2H_6O \)[/tex], so we need 3 water molecules [tex]\( 3H_2O \)[/tex] on the products side to balance the hydrogen.
[tex]\[
C_2H_6O + O_2 \rightarrow 2CO_2 + 3H_2O
\][/tex]
- Step 4: Balance the oxygen atoms.
- On the products side, we have [tex]\( 2 \times 2 = 4 \)[/tex] oxygen atoms in [tex]\( 2CO_2 \)[/tex] and [tex]\( 3 \times 1 = 3 \)[/tex] in [tex]\( 3H_2O \)[/tex], totaling [tex]\( 7 \)[/tex] oxygen atoms.
- On the reactants side, [tex]\( C_2H_6O \)[/tex] provides [tex]\( 1 \)[/tex] oxygen atom, so we need [tex]\( 6 \)[/tex] more from [tex]\( O_2 \)[/tex], hence we need [tex]\( 3 \)[/tex] molecules of [tex]\( O_2 \)[/tex].
[tex]\[
C_2H_6O + 3O_2 \rightarrow 2CO_2 + 3H_2O
\][/tex]
Hence, the balanced equation is:
[tex]\[
C_2H_6O + 3O_2 \rightarrow 2CO_2 + 3H_2O
\][/tex]
### Summary of Balanced Equations:
[tex]\[
\begin{array}{l}
CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \\
CaCl_2 + 2AgNO_3 \rightarrow Ca(NO_3)_2 + 2AgCl \\
C_2H_6O + 3O_2 \rightarrow 2CO_2 + 3H_2O
\end{array}
\][/tex]
These are the balanced versions of the given equations.
