Answer :
Let's work through balancing each chemical equation, step-by-step:
### Part a) Combustion of Decane (C₁₀H₂₂)
The unbalanced equation is:
[tex]\[ C_{10}H_{22} + O_2 \rightarrow CO_2 + H_2O \][/tex]
1. Balance Carbon (C) atoms:
[tex]\[ C_{10}H_{22} + O_2 \rightarrow 10 CO_2 + H_2O \][/tex]
2. Balance Hydrogen (H) atoms:
[tex]\[ C_{10}H_{22} + O_2 \rightarrow 10 CO_2 + 11 H_2O \][/tex]
(22 H in decane and 2 H in each water molecule → 22/2 = 11 H₂O)
3. Balance Oxygen (O) atoms:
- On the right, we have [tex]\(10 \times 2 + 11 \times 1\)[/tex] oxygens = 20 + 11 = 31 O atoms.
- Each O₂ molecule provides 2 O atoms.
[tex]\[ C_{10}H_{22} + \frac{31}{2} O_2 \rightarrow 10 CO_2 + 11 H_2O \][/tex]
For simplicity, we multiply by 2 to clear the fraction:
[tex]\[ 2 C_{10}H_{22} + 31 O_2 \rightarrow 20 CO_2 + 22 H_2O \][/tex]
So the balanced equation is:
[tex]\[\boxed{2 C_{10}H_{22} + 31 O_2 \rightarrow 20 CO_2 + 22 H_2O}\][/tex]
### Part b) Reaction of Bismuth with Nitric Acid
The unbalanced equation is:
[tex]\[ Bi + HNO_3 + H_2O \rightarrow Bi(NO_3)_3 \cdot 5 H_2O + NO \][/tex]
1. Balance Bismuth (Bi):
[tex]\[ Bi + HNO_3 + H_2O \rightarrow 1 Bi(NO_3)_3 \cdot 5 H_2O + NO \][/tex]
2. Balance the nitrate ions NO₃:
(Bi(NO₃)₃ indicates 3 nitrate ions (NO₃⁻))
[tex]\[ Bi + 3 HNO_3 + H_2O \rightarrow Bi(NO_3)_3 \cdot 5 H_2O + NO \][/tex]
3. Balance Nitrogen (N) atoms:
- There is 1 NO produced, so we need to balance the nitrogen in all other terms.
Since the number of nitrate ions remains constant:
[tex]\[ Bi + 8 HNO_3 + 5 H_2O \rightarrow 1 Bi(NO_3)_3 \cdot 5 H_2O + 3 NO \][/tex]
4. Balance Oxygen (O) and Hydrogen (H) atoms:
- Check if all H and O atoms are balanced:
- Left side: 8 nitrates = 24 O + 5 H₂O = 5 O + 6 H from HNO₃ and Bi(NO₃)₃
- Right side: [tex]\(15 O + 6 H\)[/tex] from Bi(NO₃)₅ + 3 from NO
So the balanced equation is:
[tex]\[\boxed{1 Bi + 8 HNO_3 + 5 H_2O \rightarrow 1 Bi(NO_3)_3 \cdot 5 H_2O + 3 NO}\][/tex]
### Part c) Reaction between Ammonium Chromate and Iron(III) Bromide
The unbalanced equation is:
[tex]\[ (NH_4)_2CrO_4 + FeBr_3 \rightarrow Fe_2(CrO_4)_3 + NH_4Br \][/tex]
1. Balance the Chromium (Cr) atoms:
[tex]\[ (NH_4)_2CrO_4 + FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + NH_4Br \][/tex]
(Chromium 3 as ferrous CrO₄⁻³)
2. Balance the Iron (Fe) atoms:
[tex]\[ (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + NH_4Br \][/tex]
3. Balance Bromine (Br) atoms:
[tex]\[ (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow Fe_2(CrO_4)_3 + 6 NH_4Br \][/tex]
4. Balance Ammonium (NH₄) atoms:
[tex]\[ 3 (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + 6 NH_4Br \][/tex]
So the balanced equation is:
[tex]\[\boxed{3 (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + 6 NH_4Br}\][/tex]
### Final Answers:
a) [tex]\( \boxed{2 C_{10}H_{22} + 31 O_2 \rightarrow 20 CO_2 + 22 H_2O} \)[/tex]
b) [tex]\( \boxed{1 Bi + 8 HNO_3 + 5 H_2O \rightarrow 1 Bi(NO_3)_3 \cdot 5 H_2O + 3 NO} \)[/tex]
c) [tex]\( \boxed{3 (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + 6 NH_4Br} \)[/tex]
### Part a) Combustion of Decane (C₁₀H₂₂)
The unbalanced equation is:
[tex]\[ C_{10}H_{22} + O_2 \rightarrow CO_2 + H_2O \][/tex]
1. Balance Carbon (C) atoms:
[tex]\[ C_{10}H_{22} + O_2 \rightarrow 10 CO_2 + H_2O \][/tex]
2. Balance Hydrogen (H) atoms:
[tex]\[ C_{10}H_{22} + O_2 \rightarrow 10 CO_2 + 11 H_2O \][/tex]
(22 H in decane and 2 H in each water molecule → 22/2 = 11 H₂O)
3. Balance Oxygen (O) atoms:
- On the right, we have [tex]\(10 \times 2 + 11 \times 1\)[/tex] oxygens = 20 + 11 = 31 O atoms.
- Each O₂ molecule provides 2 O atoms.
[tex]\[ C_{10}H_{22} + \frac{31}{2} O_2 \rightarrow 10 CO_2 + 11 H_2O \][/tex]
For simplicity, we multiply by 2 to clear the fraction:
[tex]\[ 2 C_{10}H_{22} + 31 O_2 \rightarrow 20 CO_2 + 22 H_2O \][/tex]
So the balanced equation is:
[tex]\[\boxed{2 C_{10}H_{22} + 31 O_2 \rightarrow 20 CO_2 + 22 H_2O}\][/tex]
### Part b) Reaction of Bismuth with Nitric Acid
The unbalanced equation is:
[tex]\[ Bi + HNO_3 + H_2O \rightarrow Bi(NO_3)_3 \cdot 5 H_2O + NO \][/tex]
1. Balance Bismuth (Bi):
[tex]\[ Bi + HNO_3 + H_2O \rightarrow 1 Bi(NO_3)_3 \cdot 5 H_2O + NO \][/tex]
2. Balance the nitrate ions NO₃:
(Bi(NO₃)₃ indicates 3 nitrate ions (NO₃⁻))
[tex]\[ Bi + 3 HNO_3 + H_2O \rightarrow Bi(NO_3)_3 \cdot 5 H_2O + NO \][/tex]
3. Balance Nitrogen (N) atoms:
- There is 1 NO produced, so we need to balance the nitrogen in all other terms.
Since the number of nitrate ions remains constant:
[tex]\[ Bi + 8 HNO_3 + 5 H_2O \rightarrow 1 Bi(NO_3)_3 \cdot 5 H_2O + 3 NO \][/tex]
4. Balance Oxygen (O) and Hydrogen (H) atoms:
- Check if all H and O atoms are balanced:
- Left side: 8 nitrates = 24 O + 5 H₂O = 5 O + 6 H from HNO₃ and Bi(NO₃)₃
- Right side: [tex]\(15 O + 6 H\)[/tex] from Bi(NO₃)₅ + 3 from NO
So the balanced equation is:
[tex]\[\boxed{1 Bi + 8 HNO_3 + 5 H_2O \rightarrow 1 Bi(NO_3)_3 \cdot 5 H_2O + 3 NO}\][/tex]
### Part c) Reaction between Ammonium Chromate and Iron(III) Bromide
The unbalanced equation is:
[tex]\[ (NH_4)_2CrO_4 + FeBr_3 \rightarrow Fe_2(CrO_4)_3 + NH_4Br \][/tex]
1. Balance the Chromium (Cr) atoms:
[tex]\[ (NH_4)_2CrO_4 + FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + NH_4Br \][/tex]
(Chromium 3 as ferrous CrO₄⁻³)
2. Balance the Iron (Fe) atoms:
[tex]\[ (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + NH_4Br \][/tex]
3. Balance Bromine (Br) atoms:
[tex]\[ (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow Fe_2(CrO_4)_3 + 6 NH_4Br \][/tex]
4. Balance Ammonium (NH₄) atoms:
[tex]\[ 3 (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + 6 NH_4Br \][/tex]
So the balanced equation is:
[tex]\[\boxed{3 (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + 6 NH_4Br}\][/tex]
### Final Answers:
a) [tex]\( \boxed{2 C_{10}H_{22} + 31 O_2 \rightarrow 20 CO_2 + 22 H_2O} \)[/tex]
b) [tex]\( \boxed{1 Bi + 8 HNO_3 + 5 H_2O \rightarrow 1 Bi(NO_3)_3 \cdot 5 H_2O + 3 NO} \)[/tex]
c) [tex]\( \boxed{3 (NH_4)_2CrO_4 + 2 FeBr_3 \rightarrow 1 Fe_2(CrO_4)_3 + 6 NH_4Br} \)[/tex]
