IDNLearn.com is designed to help you find reliable answers to any question you have. Get the information you need from our community of experts, who provide detailed and trustworthy answers.
Sagot :
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]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Your search for solutions ends here at IDNLearn.com. Thank you for visiting, and come back soon for more helpful information.
