Answer :
Sure! Let's balance each chemical equation step by step.
### Part (a)
The unbalanced equation is:
[tex]\[ \text{Al} + \text{NaOH} \rightarrow \text{NaAlO}_3 + \text{H}_2 \][/tex]
1. Aluminum (Al):
There is 1 Al atom on the left and 1 NaAlO3 on the right.
[tex]\[ \text{Al} \rightarrow \text{NaAlO}_3 \][/tex]
2. Balance sodium (Na):
There is Na in NaOH and should be consistent when forming NaAlO3. This must be factored into balancing.
[tex]\[ \text{NaOH} \rightarrow \text{NaAlO}_3 \][/tex]
3. Balance Hydrogen (H):
There is H in NaOH and H2. We should balance the number of hydrogen atoms.
[tex]\[ \text{NaOH} + \text{Al} \rightarrow \text{NaAlO}_3 + \text{H}_2 \][/tex]
After analyzing the elements:
- Aluminum and sodium proportion match.
To balance this equation:
- We find the coefficients to ensure the number of each type of atom on both sides matches:
[tex]\[ Al + 3NaOH \rightarrow NaAlO_3 + 1.5H_2 \][/tex]
This simplifies to:
[tex]\[ 2Al + 6NaOH \rightarrow 2NaAlO_3 + 3H_2 \][/tex]
### Part (b)
The unbalanced equation is:
[tex]\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \][/tex]
1. Carbon (C):
There are 12 C atoms on the left (from C12H22O11), so we need 12 CO2 on the right.
[tex]\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} \rightarrow 12\text{CO}_2 \][/tex]
2. Hydrogen (H):
There are 22 H atoms on the left, so we need 11 H2O on the right.
[tex]\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} \rightarrow 12\text{CO}_2 + 11\text{H}_2\text{O} \][/tex]
3. Oxygen (O):
There are 11 O atoms in C12H22O11 and 24 from 12 CO2 and 11 H2O totals:
On the left: [tex]\(11 + X (O_2)\)[/tex],
On the right: [tex]\(24 + 11 = 35 O\)[/tex]
Therefore, we need [tex]\(12O_2 + 11 = 23 O\)[/tex].
Thus, balancing everything:
[tex]\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \frac{23}{2}\text{O}_2 \rightarrow 12\text{CO}_2 + 11\text{H}_2\text{O} \][/tex]
### Part (c)
The unbalanced equation is:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_3\text{PO}_4 \][/tex]
1. Calcium (Ca):
There are 3 Ca on the left in Ca3(PO4)2, so we need 3 CaSO4 on the right:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 \rightarrow 3\text{CaSO}_4 \][/tex]
2. Phosphorus (P):
There are 2 PO4 on the left, so we need 2 H3PO4 on the right:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 \rightarrow 3\text{CaSO}_4 + 2\text{H}_3\text{PO}_4 \][/tex]
3. Balance Hydrogen (H):
To balance H atoms, ensuring [tex]\(3H_2SO_4\)[/tex] accounts for all S and H balances.
Thus, the balanced equation is:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 + 3\text{H}_2\text{SO}_4 \rightarrow 3\text{CaSO}_4 + 2\text{H}_3\text{PO}_4 \][/tex]
Therefore, the overall balanced equations are:
- (a): [tex]\(2 Al + 6 NaOH \rightarrow 2 NaAlO_3 + 3 H_2\)[/tex]
- (b): [tex]\(C_{12}H_{22}O_{11} + \frac{23}{2} O_2 \rightarrow 12 CO_2 + 11 H_2O\)[/tex]
- (c): [tex]\(Ca_3(PO_4)_2 + 3 H_2SO_4 \rightarrow 3 CaSO_4 + 2 H_3PO_4\)[/tex]
### Part (a)
The unbalanced equation is:
[tex]\[ \text{Al} + \text{NaOH} \rightarrow \text{NaAlO}_3 + \text{H}_2 \][/tex]
1. Aluminum (Al):
There is 1 Al atom on the left and 1 NaAlO3 on the right.
[tex]\[ \text{Al} \rightarrow \text{NaAlO}_3 \][/tex]
2. Balance sodium (Na):
There is Na in NaOH and should be consistent when forming NaAlO3. This must be factored into balancing.
[tex]\[ \text{NaOH} \rightarrow \text{NaAlO}_3 \][/tex]
3. Balance Hydrogen (H):
There is H in NaOH and H2. We should balance the number of hydrogen atoms.
[tex]\[ \text{NaOH} + \text{Al} \rightarrow \text{NaAlO}_3 + \text{H}_2 \][/tex]
After analyzing the elements:
- Aluminum and sodium proportion match.
To balance this equation:
- We find the coefficients to ensure the number of each type of atom on both sides matches:
[tex]\[ Al + 3NaOH \rightarrow NaAlO_3 + 1.5H_2 \][/tex]
This simplifies to:
[tex]\[ 2Al + 6NaOH \rightarrow 2NaAlO_3 + 3H_2 \][/tex]
### Part (b)
The unbalanced equation is:
[tex]\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \][/tex]
1. Carbon (C):
There are 12 C atoms on the left (from C12H22O11), so we need 12 CO2 on the right.
[tex]\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} \rightarrow 12\text{CO}_2 \][/tex]
2. Hydrogen (H):
There are 22 H atoms on the left, so we need 11 H2O on the right.
[tex]\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} \rightarrow 12\text{CO}_2 + 11\text{H}_2\text{O} \][/tex]
3. Oxygen (O):
There are 11 O atoms in C12H22O11 and 24 from 12 CO2 and 11 H2O totals:
On the left: [tex]\(11 + X (O_2)\)[/tex],
On the right: [tex]\(24 + 11 = 35 O\)[/tex]
Therefore, we need [tex]\(12O_2 + 11 = 23 O\)[/tex].
Thus, balancing everything:
[tex]\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \frac{23}{2}\text{O}_2 \rightarrow 12\text{CO}_2 + 11\text{H}_2\text{O} \][/tex]
### Part (c)
The unbalanced equation is:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_3\text{PO}_4 \][/tex]
1. Calcium (Ca):
There are 3 Ca on the left in Ca3(PO4)2, so we need 3 CaSO4 on the right:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 \rightarrow 3\text{CaSO}_4 \][/tex]
2. Phosphorus (P):
There are 2 PO4 on the left, so we need 2 H3PO4 on the right:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 \rightarrow 3\text{CaSO}_4 + 2\text{H}_3\text{PO}_4 \][/tex]
3. Balance Hydrogen (H):
To balance H atoms, ensuring [tex]\(3H_2SO_4\)[/tex] accounts for all S and H balances.
Thus, the balanced equation is:
[tex]\[ \text{Ca}_3(\text{PO}_4)_2 + 3\text{H}_2\text{SO}_4 \rightarrow 3\text{CaSO}_4 + 2\text{H}_3\text{PO}_4 \][/tex]
Therefore, the overall balanced equations are:
- (a): [tex]\(2 Al + 6 NaOH \rightarrow 2 NaAlO_3 + 3 H_2\)[/tex]
- (b): [tex]\(C_{12}H_{22}O_{11} + \frac{23}{2} O_2 \rightarrow 12 CO_2 + 11 H_2O\)[/tex]
- (c): [tex]\(Ca_3(PO_4)_2 + 3 H_2SO_4 \rightarrow 3 CaSO_4 + 2 H_3PO_4\)[/tex]
