Answer :
Let's balance the given chemical reactions step-by-step.
### Part (a):
Reaction: [tex]\( \_ Cu_2O(s) + \_ C(s) \rightarrow \_ Cu(s) + \_ CO(s) \)[/tex]
1. Copper (Cu):
- On the left-hand side (LHS), we have [tex]\( Cu_2O \)[/tex] which contains 2 Cu atoms.
- On the right-hand side (RHS), we need 2 Cu atoms.
- Therefore, the coefficient for [tex]\( Cu \)[/tex] should be 2.
Updated equation:
[tex]\[ Cu_2O(s) + C(s) \rightarrow 2 Cu(s) + CO(s) \][/tex]
2. Oxygen (O):
- On the LHS, [tex]\( Cu_2O \)[/tex] has 1 oxygen atom.
- On the RHS, [tex]\( CO \)[/tex] has 1 oxygen atom.
- Since oxygen is balanced (1 O on each side), we do not need to change anything for oxygen.
3. Carbon (C):
- On the LHS, we have 1 carbon atom from [tex]\( C \)[/tex].
- On the RHS, [tex]\( CO \)[/tex] has 1 carbon atom.
- Since carbon is also balanced, no further adjustments are required.
The balanced equation for part (a) is:
[tex]\[ 1 Cu_2O(s) + 1 C(s) \rightarrow 2 Cu(s) + 1 CO(s) \][/tex]
### Part (c):
Reaction: [tex]\( C_3H_8 + \_ O_2 \rightarrow \_ CO_2 + \_ H_2O \)[/tex]
1. Carbon (C):
- On the LHS, [tex]\( C_3H_8 \)[/tex] has 3 carbon atoms.
- On the RHS, we need 3 [tex]\( CO_2 \)[/tex] molecules to balance the carbon atoms.
Updated equation:
[tex]\[ C_3H_8 + O_2 \rightarrow 3 CO_2 + H_2O \][/tex]
2. Hydrogen (H):
- On the LHS, [tex]\( C_3H_8 \)[/tex] has 8 hydrogen atoms.
- On the RHS, to have 8 hydrogen atoms, we need 4 [tex]\( H_2O \)[/tex] molecules (each [tex]\( H_2O \)[/tex] has 2 hydrogen atoms).
Updated equation:
[tex]\[ C_3H_8 + O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]
3. Oxygen (O):
- On the RHS, we have oxygen atoms from [tex]\( CO_2 \)[/tex] and [tex]\( H_2O \)[/tex]:
- [tex]\( 3 CO_2 \)[/tex]: [tex]\( 3 \times 2 = 6 \)[/tex] oxygen atoms.
- [tex]\( 4 H_2O \)[/tex]: [tex]\( 4 \times 1 = 4 \)[/tex] oxygen atoms.
- Total: [tex]\( 6 + 4 = 10 \)[/tex] oxygen atoms.
- On the LHS, [tex]\( O_2 \)[/tex] has 2 oxygen atoms per molecule, so we need 5 [tex]\( O_2 \)[/tex] molecules to get 10 oxygen atoms.
The balanced equation for part (c) is:
[tex]\[ 1 C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]
### Part (d):
Reaction: [tex]\( Al(NO_3)_3 + \_ NaOH \rightarrow \_ Al(OH)_3 + \_ NaNO_3 \)[/tex]
1. Aluminum (Al):
- On the LHS, we have 1 Al atom in [tex]\( Al(NO_3)_3 \)[/tex].
- On the RHS, we have 1 Al atom in [tex]\( Al(OH)_3 \)[/tex].
- No adjustment needed for aluminum.
2. Nitrate (NO_3):
- On the LHS, [tex]\( Al(NO_3)_3 \)[/tex] has 3 nitrate ions ([tex]\( 3 NO_3 \)[/tex]).
- On the RHS, we need 3 [tex]\( NaNO_3 \)[/tex] to balance the nitrate ions.
Updated equation:
[tex]\[ Al(NO_3)_3 + NaOH \rightarrow Al(OH)_3 + 3 NaNO_3 \][/tex]
3. Sodium (Na):
- On the LHS, to balance 3 [tex]\( NaNO_3 \)[/tex] on the RHS, we need 3 [tex]\( NaOH \)[/tex] molecules (since each [tex]\( NaOH \)[/tex] contains 1 Na).
Updated equation:
[tex]\[ Al(NO_3)_3 + 3 NaOH \rightarrow Al(OH)_3 + 3 NaNO_3 \][/tex]
4. Hydroxide (OH):
- On the LHS, we have 3 [tex]\( OH \)[/tex] ions from 3 [tex]\( NaOH \)[/tex].
- On the RHS, we need 3 [tex]\( OH \)[/tex] ions in [tex]\( Al(OH)_3 \)[/tex] to balance the hydroxide.
The balanced equation for part (d) is:
[tex]\[ 1 Al(NO_3)_3 + 3 NaOH \rightarrow 1 Al(OH)_3 + 3 NaNO_3 \][/tex]
Thus, the balanced chemical equations are:
a. [tex]\( 1 Cu_2O(s) + 1 C(s) \rightarrow 2 Cu(s) + 1 CO(s) \)[/tex]
c. [tex]\( 1 C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \)[/tex]
d. [tex]\( 1 Al(NO_3)_3 + 3 NaOH \rightarrow 1 Al(OH)_3 + 3 NaNO_3 \)[/tex]
### Part (a):
Reaction: [tex]\( \_ Cu_2O(s) + \_ C(s) \rightarrow \_ Cu(s) + \_ CO(s) \)[/tex]
1. Copper (Cu):
- On the left-hand side (LHS), we have [tex]\( Cu_2O \)[/tex] which contains 2 Cu atoms.
- On the right-hand side (RHS), we need 2 Cu atoms.
- Therefore, the coefficient for [tex]\( Cu \)[/tex] should be 2.
Updated equation:
[tex]\[ Cu_2O(s) + C(s) \rightarrow 2 Cu(s) + CO(s) \][/tex]
2. Oxygen (O):
- On the LHS, [tex]\( Cu_2O \)[/tex] has 1 oxygen atom.
- On the RHS, [tex]\( CO \)[/tex] has 1 oxygen atom.
- Since oxygen is balanced (1 O on each side), we do not need to change anything for oxygen.
3. Carbon (C):
- On the LHS, we have 1 carbon atom from [tex]\( C \)[/tex].
- On the RHS, [tex]\( CO \)[/tex] has 1 carbon atom.
- Since carbon is also balanced, no further adjustments are required.
The balanced equation for part (a) is:
[tex]\[ 1 Cu_2O(s) + 1 C(s) \rightarrow 2 Cu(s) + 1 CO(s) \][/tex]
### Part (c):
Reaction: [tex]\( C_3H_8 + \_ O_2 \rightarrow \_ CO_2 + \_ H_2O \)[/tex]
1. Carbon (C):
- On the LHS, [tex]\( C_3H_8 \)[/tex] has 3 carbon atoms.
- On the RHS, we need 3 [tex]\( CO_2 \)[/tex] molecules to balance the carbon atoms.
Updated equation:
[tex]\[ C_3H_8 + O_2 \rightarrow 3 CO_2 + H_2O \][/tex]
2. Hydrogen (H):
- On the LHS, [tex]\( C_3H_8 \)[/tex] has 8 hydrogen atoms.
- On the RHS, to have 8 hydrogen atoms, we need 4 [tex]\( H_2O \)[/tex] molecules (each [tex]\( H_2O \)[/tex] has 2 hydrogen atoms).
Updated equation:
[tex]\[ C_3H_8 + O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]
3. Oxygen (O):
- On the RHS, we have oxygen atoms from [tex]\( CO_2 \)[/tex] and [tex]\( H_2O \)[/tex]:
- [tex]\( 3 CO_2 \)[/tex]: [tex]\( 3 \times 2 = 6 \)[/tex] oxygen atoms.
- [tex]\( 4 H_2O \)[/tex]: [tex]\( 4 \times 1 = 4 \)[/tex] oxygen atoms.
- Total: [tex]\( 6 + 4 = 10 \)[/tex] oxygen atoms.
- On the LHS, [tex]\( O_2 \)[/tex] has 2 oxygen atoms per molecule, so we need 5 [tex]\( O_2 \)[/tex] molecules to get 10 oxygen atoms.
The balanced equation for part (c) is:
[tex]\[ 1 C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \][/tex]
### Part (d):
Reaction: [tex]\( Al(NO_3)_3 + \_ NaOH \rightarrow \_ Al(OH)_3 + \_ NaNO_3 \)[/tex]
1. Aluminum (Al):
- On the LHS, we have 1 Al atom in [tex]\( Al(NO_3)_3 \)[/tex].
- On the RHS, we have 1 Al atom in [tex]\( Al(OH)_3 \)[/tex].
- No adjustment needed for aluminum.
2. Nitrate (NO_3):
- On the LHS, [tex]\( Al(NO_3)_3 \)[/tex] has 3 nitrate ions ([tex]\( 3 NO_3 \)[/tex]).
- On the RHS, we need 3 [tex]\( NaNO_3 \)[/tex] to balance the nitrate ions.
Updated equation:
[tex]\[ Al(NO_3)_3 + NaOH \rightarrow Al(OH)_3 + 3 NaNO_3 \][/tex]
3. Sodium (Na):
- On the LHS, to balance 3 [tex]\( NaNO_3 \)[/tex] on the RHS, we need 3 [tex]\( NaOH \)[/tex] molecules (since each [tex]\( NaOH \)[/tex] contains 1 Na).
Updated equation:
[tex]\[ Al(NO_3)_3 + 3 NaOH \rightarrow Al(OH)_3 + 3 NaNO_3 \][/tex]
4. Hydroxide (OH):
- On the LHS, we have 3 [tex]\( OH \)[/tex] ions from 3 [tex]\( NaOH \)[/tex].
- On the RHS, we need 3 [tex]\( OH \)[/tex] ions in [tex]\( Al(OH)_3 \)[/tex] to balance the hydroxide.
The balanced equation for part (d) is:
[tex]\[ 1 Al(NO_3)_3 + 3 NaOH \rightarrow 1 Al(OH)_3 + 3 NaNO_3 \][/tex]
Thus, the balanced chemical equations are:
a. [tex]\( 1 Cu_2O(s) + 1 C(s) \rightarrow 2 Cu(s) + 1 CO(s) \)[/tex]
c. [tex]\( 1 C_3H_8 + 5 O_2 \rightarrow 3 CO_2 + 4 H_2O \)[/tex]
d. [tex]\( 1 Al(NO_3)_3 + 3 NaOH \rightarrow 1 Al(OH)_3 + 3 NaNO_3 \)[/tex]
