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To balance the given chemical equations step-by-step, let's proceed with each reaction separately:
### 1. Balancing [tex]\( \mathrm{HCl (aq) + MgCO_3 (aq) \rightarrow MgCl_2 (aq) + H_2O (l) + CO_2 (g)} \)[/tex]
#### Step-by-Step:
1. Identify the number of atoms of each element on both sides:
- Reactants side:
- H: 1 (from HCl)
- Cl: 1 (from HCl)
- Mg: 1 (from MgCO[tex]\(_3\)[/tex])
- C: 1 (from MgCO[tex]\(_3\)[/tex])
- O: 3 (from MgCO[tex]\(_3\)[/tex])
- Products side:
- Mg: 1 (from MgCl[tex]\(_2\)[/tex])
- Cl: 2 (from MgCl[tex]\(_2\)[/tex])
- H: 2 (from H[tex]\(_2\)[/tex]O)
- O: 3 (2 from CO[tex]\(_2\)[/tex] and 1 from H[tex]\(_2\)[/tex]O)
- C: 1 (from CO[tex]\(_2\)[/tex])
2. Balance Chlorine (Cl):
- We have 1 Cl on the reactants side and 2 Cl on the products side.
- Multiply HCl by 2 on the reactants side:
[tex]\[ 2 \mathrm{HCl} + \mathrm{MgCO}_3 \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \][/tex]
3. Check other atoms:
- H: 2 (reactants) and 2 (products) – balanced
- Mg: 1 (reactants) and 1 (products) – balanced
- C: 1 (reactants) and 1 (products) – balanced
- O: 3 (reactants) and 3 (products) – balanced
The balanced equation is:
[tex]\[ 2 \mathrm{HCl} + \mathrm{MgCO}_3 \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \][/tex]
### 2. Balancing [tex]\( \mathrm{C_4H_{10} (g) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g)} \)[/tex]
#### Step-by-Step:
1. Identify the number of atoms of each element on both sides:
- Reactants side:
- C: 4 (from C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex])
- H: 10 (from C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex])
- O: 2 (from O[tex]\(_2\)[/tex])
- Products side:
- C: 1 (from CO[tex]\(_2\)[/tex])
- H: 2 (from H[tex]\(_2\)[/tex]O)
- O: 3 (2 from CO[tex]\(_2\)[/tex] and 1 from H[tex]\(_2\)[/tex]O)
2. Balance Carbon (C):
- We have 4 C on the reactants side and 1 C on the products side.
- Multiply CO[tex]\(_2\)[/tex] by 4:
[tex]\[ \mathrm{C}_4\mathrm{H}_{10} + \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \][/tex]
3. Balance Hydrogen (H):
- We have 10 H on the reactants side and 2 H on the products side.
- Multiply H[tex]\(_2\)[/tex]O by 5:
[tex]\[ \mathrm{C}_4\mathrm{H}_{10} + \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 5 \mathrm{H}_2\mathrm{O} \][/tex]
4. Balance Oxygen (O):
- On the products side, we now have 4 CO[tex]\(_2\)[/tex] which give us [tex]\(4 \times 2 = 8\)[/tex] O, plus 5 H[tex]\(_2\)[/tex]O that give us [tex]\(5 \times 1 = 5\)[/tex] O. Total: [tex]\(8 + 5 = 13\)[/tex].
- On the reactants side, oxygen needs to be balanced by 13 O. Therefore, we need [tex]\( \frac{13}{2} \)[/tex] O[tex]\(_2\)[/tex] molecules.
- Multiply the entire equation by 2 to clear the fraction:
[tex]\[ 2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O} \][/tex]
The balanced equation is:
[tex]\[ 2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O} \][/tex]
### 3. Balancing [tex]\( \mathrm{Al (s) + O_2 (g) \rightarrow Al_2O_3 (s)} \)[/tex]
#### Step-by-Step:
1. Identify the number of atoms of each element on both sides:
- Reactants side:
- Al: 1 (from Al)
- O: 2 (from O[tex]\(_2\)[/tex])
- Products side:
- Al: 2 (from Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex])
- O: 3 (from Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex])
2. Balance Aluminum (Al):
- We have 1 Al on the reactants side and 2 Al on the products side.
- Multiply Al by 2:
[tex]\[ 2 \mathrm{Al} + \mathrm{O}_2 \rightarrow \mathrm{Al}_2\mathrm{O}_3 \][/tex]
3. Balance Oxygen (O):
- We have 2 O on the reactants side and 3 O on the products side.
- Multiply O[tex]\(_2\)[/tex] by [tex]\( \frac{3}{2} \)[/tex]:
[tex]\[ 2 \mathrm{Al} + \frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{Al}_2\mathrm{O}_3 \][/tex]
- Multiply the entire equation by 2 to clear the fraction:
[tex]\[ 4 \mathrm{Al} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2\mathrm{O}_3 \][/tex]
The balanced equation is:
[tex]\[ 4 \mathrm{Al} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2\mathrm{O}_3 \][/tex]
These are the step-by-step solutions for balancing the given chemical equations:
1. [tex]\(2 \mathrm{HCl} + \mathrm{MgCO}_3 \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\)[/tex]
2. [tex]\(2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O}\)[/tex]
3. [tex]\(4 \mathrm{Al} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2\mathrm{O}_3\)[/tex]
### 1. Balancing [tex]\( \mathrm{HCl (aq) + MgCO_3 (aq) \rightarrow MgCl_2 (aq) + H_2O (l) + CO_2 (g)} \)[/tex]
#### Step-by-Step:
1. Identify the number of atoms of each element on both sides:
- Reactants side:
- H: 1 (from HCl)
- Cl: 1 (from HCl)
- Mg: 1 (from MgCO[tex]\(_3\)[/tex])
- C: 1 (from MgCO[tex]\(_3\)[/tex])
- O: 3 (from MgCO[tex]\(_3\)[/tex])
- Products side:
- Mg: 1 (from MgCl[tex]\(_2\)[/tex])
- Cl: 2 (from MgCl[tex]\(_2\)[/tex])
- H: 2 (from H[tex]\(_2\)[/tex]O)
- O: 3 (2 from CO[tex]\(_2\)[/tex] and 1 from H[tex]\(_2\)[/tex]O)
- C: 1 (from CO[tex]\(_2\)[/tex])
2. Balance Chlorine (Cl):
- We have 1 Cl on the reactants side and 2 Cl on the products side.
- Multiply HCl by 2 on the reactants side:
[tex]\[ 2 \mathrm{HCl} + \mathrm{MgCO}_3 \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \][/tex]
3. Check other atoms:
- H: 2 (reactants) and 2 (products) – balanced
- Mg: 1 (reactants) and 1 (products) – balanced
- C: 1 (reactants) and 1 (products) – balanced
- O: 3 (reactants) and 3 (products) – balanced
The balanced equation is:
[tex]\[ 2 \mathrm{HCl} + \mathrm{MgCO}_3 \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2 \][/tex]
### 2. Balancing [tex]\( \mathrm{C_4H_{10} (g) + O_2 (g) \rightarrow CO_2 (g) + H_2O (g)} \)[/tex]
#### Step-by-Step:
1. Identify the number of atoms of each element on both sides:
- Reactants side:
- C: 4 (from C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex])
- H: 10 (from C[tex]\(_4\)[/tex]H[tex]\(_{10}\)[/tex])
- O: 2 (from O[tex]\(_2\)[/tex])
- Products side:
- C: 1 (from CO[tex]\(_2\)[/tex])
- H: 2 (from H[tex]\(_2\)[/tex]O)
- O: 3 (2 from CO[tex]\(_2\)[/tex] and 1 from H[tex]\(_2\)[/tex]O)
2. Balance Carbon (C):
- We have 4 C on the reactants side and 1 C on the products side.
- Multiply CO[tex]\(_2\)[/tex] by 4:
[tex]\[ \mathrm{C}_4\mathrm{H}_{10} + \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O} \][/tex]
3. Balance Hydrogen (H):
- We have 10 H on the reactants side and 2 H on the products side.
- Multiply H[tex]\(_2\)[/tex]O by 5:
[tex]\[ \mathrm{C}_4\mathrm{H}_{10} + \mathrm{O}_2 \rightarrow 4 \mathrm{CO}_2 + 5 \mathrm{H}_2\mathrm{O} \][/tex]
4. Balance Oxygen (O):
- On the products side, we now have 4 CO[tex]\(_2\)[/tex] which give us [tex]\(4 \times 2 = 8\)[/tex] O, plus 5 H[tex]\(_2\)[/tex]O that give us [tex]\(5 \times 1 = 5\)[/tex] O. Total: [tex]\(8 + 5 = 13\)[/tex].
- On the reactants side, oxygen needs to be balanced by 13 O. Therefore, we need [tex]\( \frac{13}{2} \)[/tex] O[tex]\(_2\)[/tex] molecules.
- Multiply the entire equation by 2 to clear the fraction:
[tex]\[ 2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O} \][/tex]
The balanced equation is:
[tex]\[ 2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O} \][/tex]
### 3. Balancing [tex]\( \mathrm{Al (s) + O_2 (g) \rightarrow Al_2O_3 (s)} \)[/tex]
#### Step-by-Step:
1. Identify the number of atoms of each element on both sides:
- Reactants side:
- Al: 1 (from Al)
- O: 2 (from O[tex]\(_2\)[/tex])
- Products side:
- Al: 2 (from Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex])
- O: 3 (from Al[tex]\(_2\)[/tex]O[tex]\(_3\)[/tex])
2. Balance Aluminum (Al):
- We have 1 Al on the reactants side and 2 Al on the products side.
- Multiply Al by 2:
[tex]\[ 2 \mathrm{Al} + \mathrm{O}_2 \rightarrow \mathrm{Al}_2\mathrm{O}_3 \][/tex]
3. Balance Oxygen (O):
- We have 2 O on the reactants side and 3 O on the products side.
- Multiply O[tex]\(_2\)[/tex] by [tex]\( \frac{3}{2} \)[/tex]:
[tex]\[ 2 \mathrm{Al} + \frac{3}{2} \mathrm{O}_2 \rightarrow \mathrm{Al}_2\mathrm{O}_3 \][/tex]
- Multiply the entire equation by 2 to clear the fraction:
[tex]\[ 4 \mathrm{Al} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2\mathrm{O}_3 \][/tex]
The balanced equation is:
[tex]\[ 4 \mathrm{Al} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2\mathrm{O}_3 \][/tex]
These are the step-by-step solutions for balancing the given chemical equations:
1. [tex]\(2 \mathrm{HCl} + \mathrm{MgCO}_3 \rightarrow \mathrm{MgCl}_2 + \mathrm{H}_2\mathrm{O} + \mathrm{CO}_2\)[/tex]
2. [tex]\(2 \mathrm{C}_4\mathrm{H}_{10} + 13 \mathrm{O}_2 \rightarrow 8 \mathrm{CO}_2 + 10 \mathrm{H}_2\mathrm{O}\)[/tex]
3. [tex]\(4 \mathrm{Al} + 3 \mathrm{O}_2 \rightarrow 2 \mathrm{Al}_2\mathrm{O}_3\)[/tex]
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