balance the following equations by placing small whole - number coefficients in the appropriate places.
synthesis
1. ba + o₂ → bao
2. c + s → cs₂
3. li + o₂ → li₂o
4. mg + n₂ → mg₃n₂
5. fecl₂ + cl₂ → fecl₃
decomposition
6. kclo₃ → kcl + o₂
7. ag₂o → ag + o₂
8. cuco₃ → cuo + co₂
9. aubr₃ → au + br₂
10. lif₄ → li + f₂
single replacement
11. zn + hcl → zncl₂ + h₂
12. zn + cuso₄ → znso₄ + cu
13. cu + agno₃ → cu(no₃)₂ + ag
14. k + h₂o → koh + h₂
15. al + cucl₂ → alcl₃ + cu
double replacement
16. bacl₂ + na₂so₄ → nacl + baso₄
17. zncl₂ + (nh₄)₂s → nh₄cl + zns
18. naoh + hcl → nacl + h₂o
19. fes + hcl → fecl₂ + h₂s
20. alcl₃ + naoh →
combustion
21. c₂h₆ + o₂ → co₂ + h₂o
22. c₄h₁₀ + o₂ → co₂ + h₂o
23. c₂h₆o + o₂ → co₂ + h₂o

Question

Accepted Answer

Let's take problem 1 (Synthesis: $\ce{Ba + O_{2} -> BaO}$) as an example to balance the equation.

# Explanation:
## Step1: Count atoms on each side
Left side: Ba = 1, O = 2; Right side: Ba = 1, O = 1.
## Step2: Balance O atoms
To balance O, multiply $\ce{BaO}$ by 2: $\ce{Ba + O_{2} -> 2BaO}$. Now O is balanced (2 on each side), but Ba is 1 on left and 2 on right.
## Step3: Balance Ba atoms
Multiply $\ce{Ba}$ by 2: $\ce{2Ba + O_{2} -> 2BaO}$. Now Ba: 2 on left and 2 on right; O: 2 on left and 2 on right.
# Answer:
$\ce{2Ba + O_{2} -> 2BaO}$

For problem 2 (Synthesis: $\ce{C + S -> CS_{2}}$):
# Explanation:
## Step1: Count atoms
Left: C = 1, S = 1; Right: C = 1, S = 2.
## Step2: Balance S
Multiply S by 2: $\ce{C + 2S -> CS_{2}}$. Now C:1, S:2 on both sides.
# Answer:
$\ce{C + 2S -> CS_{2}}$

For problem 3 (Synthesis: $\ce{Li + O_{2} -> Li_{2}O}$):
# Explanation:
## Step1: Count atoms
Left: Li = 1, O = 2; Right: Li = 2, O = 1.
## Step2: Balance O
Multiply $\ce{Li_{2}O}$ by 2: $\ce{Li + O_{2} -> 2Li_{2}O}$. Now O:2 on both sides, Li:1 left, 4 right.
## Step3: Balance Li
Multiply Li by 4: $\ce{4Li + O_{2} -> 2Li_{2}O}$. Now Li:4, O:2 on both sides. Wait, but we can simplify. Wait, better way:
Start over. Let's balance O first. O on left: 2, right:1. So multiply $\ce{Li_{2}O}$ by 2 (O:2). Then Li on right: 4, so multiply Li by 4? No, wait, maybe a better approach. Wait, the correct way:
Left: O = 2, right: O = 1 (in $\ce{Li_{2}O}$). So to balance O, we need 2 O on right, so $\ce{Li_{2}O}$ * 2 (O:2). Then Li: 4 on right, so Li * 4? But that gives 4 Li + O₂ -> 2 Li₂O. Let's check: Li:4, O:2 on both sides. Yes.
But alternatively, maybe I made a mistake. Wait, the correct balanced equation is $\ce{4Li + O_{2} -> 2Li_{2}O}$.
# Answer:
$\ce{4Li + O_{2} -> 2Li_{2}O}$

For problem 4 (Synthesis: $\ce{Mg + N_{2} -> Mg_{3}N_{2}}$):
# Explanation:
## Step1: Count atoms
Left: Mg = 1, N = 2; Right: Mg = 3, N = 2.
## Step2: Balance Mg
Multiply Mg by 3: $\ce{3Mg + N_{2} -> Mg_{3}N_{2}}$. Now Mg:3, N:2 on both sides.
# Answer:
$\ce{3Mg + N_{2} -> Mg_{3}N_{2}}$

For problem 5 (Synthesis: $\ce{FeCl_{2} + Cl_{2} -> FeCl_{3}}$):
# Explanation:
## Step1: Count atoms
Left: Fe = 1, Cl = 2 + 2 = 4; Right: Fe = 1, Cl = 3.
## Step2: Balance Cl
Find least common multiple of 4 and 3, which is 12? No, better: Let's see. Left Cl: 4, Right Cl:3. Multiply $\ce{FeCl_{2}}$ by 2: $\ce{2FeCl_{2} + Cl_{2} -> FeCl_{3}}$. Now Fe:2, Cl:4 + 2 = 6 on left. Right Cl:3, so multiply $\ce{FeCl_{3}}$ by 2: $\ce{2FeCl_{2} + Cl_{2} -> 2FeCl_{3}}$. Now Fe:2, Cl:4 + 2 = 6 on left; Fe:2, Cl:6 on right. Balanced.
# Answer:
$\ce{2FeCl_{2} + Cl_{2} -> 2FeCl_{3}}$

For problem 6 (Decomposition: $\ce{KClO_{3} -> KCl + O_{2}}$):
# Explanation:
## Step1: Count atoms
Left: K = 1, Cl = 1, O = 3; Right: K = 1, Cl = 1, O = 2.
## Step2: Balance O
Multiply $\ce{KClO_{3}}$ by 2 and $\ce{O_{2}}$ by 3: $\ce{2KClO_{3} -> KCl + 3O_{2}}$. Now O:6 on both sides. K and Cl: 2 on left, 1 on right.
## Step3: Balance K and Cl
Multiply $\ce{KCl}$ by 2: $\ce{2KClO_{3} -> 2KCl + 3O_{2}}$. Now K:2, Cl:2, O:6 on both sides.
# Answer:
$\ce{2KClO_{3} -> 2KCl + 3O_{2}}$ (with appropriate conditions like heating and catalyst, but for balancing, this is the equation)

For problem 7 (Decomposition: $\ce{Ag_{2}O -> Ag + O_{2}}$):
# Explanation:
## Step1: Count atoms
Left: Ag = 2, O = 1; Right: Ag = 1, O = 2.
## Step2: Balance O
Multiply $\ce{Ag_{2}O}$ by 2: $\ce{2Ag_{2}O -> Ag + O_{2}}$. Now O:2 on both sides. Ag:4 on left, 1 on right.
## Step3: Balance Ag
Multiply Ag by 4: $\ce{2Ag_{2}O -> 4Ag + O_{2}}$. Now Ag:4, O:2 on both sides.
# Answer:
$\ce{2Ag_{2}O -> 4Ag + O_{2}}$

For problem 8 (Decomposition: $\ce{CuCO_{3} -> CuO + CO_{2}}$):
# Explanation:
## Step1: Count atoms
Left: Cu = 1, C = 1, O = 3; Right: Cu = 1, C = 1, O = 1 + 2 = 3. Already balanced.
# Answer:
$\ce{CuCO_{3} -> CuO + CO_{2}}$ (already balanced)

For problem 9 (Decomposition: $\ce{AuBr_{3} -> Au + Br_{2}}$):
# Explanation:
## Step1: Count atoms
Left: Au = 1, Br = 3; Right: Au = 1, Br = 2.
## Step2: Balance Br
Multiply $\ce{AuBr_{3}}$ by 2 and $\ce{Br_{2}}$ by 3: $\ce{2AuBr_{3} -> Au + 3Br_{2}}$. Now Br:6 on both sides. Au:2 on left, 1 on right.
## Step3: Balance Au
Multiply Au by 2: $\ce{2AuBr_{3} -> 2Au + 3Br_{2}}$. Now Au:2, Br:6 on both sides.
# Answer:
$\ce{2AuBr_{3} -> 2Au + 3Br_{2}}$

For problem 10 (Decomposition: $\ce{LiF_{4} -> Li + F_{2}}$) – Wait, probably a typo, should be $\ce{LiF}$? No, maybe $\ce{LiF_{2}}$? Wait, original is $\ce{LiF_{4}}$? Wait, maybe it's $\ce{LiF}$ or $\ce{LiF_{2}}$. Wait, assuming it's $\ce{LiF_{2}}$ (maybe a typo, but let's proceed with $\ce{LiF_{2}}$: $\ce{LiF_{2} -> Li + F_{2}}$? No, left F:2, right F:2. Li:1 on both. Wait, no, original is $\ce{LiF_{4}}$: left Li:1, F:4; right Li:1, F:2. So balance F: multiply $\ce{F_{2}}$ by 2: $\ce{LiF_{4} -> Li + 2F_{2}}$. Now Li:1, F:4 on both sides. So:
# Explanation:
## Step1: Count atoms
Left: Li = 1, F = 4; Right: Li = 1, F = 2.
## Step2: Balance F
Multiply $\ce{F_{2}}$ by 2: $\ce{LiF_{4} -> Li + 2F_{2}}$. Now F:4 on both sides, Li:1 on both.
# Answer:
$\ce{LiF_{4} -> Li + 2F_{2}}$ (assuming the formula is correct as given)

For problem 11 (Single Replacement: $\ce{Zn + HCl -> ZnCl_{2} + H_{2}}$):
# Explanation:
## Step1: Count atoms
Left: Zn = 1, H = 1, Cl = 1; Right: Zn = 1, H = 2, Cl = 2.
## Step2: Balance Cl and H
Multiply HCl by 2: $\ce{Zn + 2HCl -> ZnCl_{2} + H_{2}}$. Now Zn:1, H:2, Cl:2 on both sides.
# Answer:
$\ce{Zn + 2HCl -> ZnCl_{2} + H_{2}}$

For problem 12 (Single Replacement: $\ce{Zn + CuSO_{4} -> ZnSO_{4} + Cu}$):
# Explanation:
## Step1: Count atoms
Left: Zn = 1, Cu = 1, S = 1, O = 4; Right: Zn = 1, Cu = 1, S = 1, O = 4. Already balanced.
# Answer:
$\ce{Zn + CuSO_{4} -> ZnSO_{4} + Cu}$ (already balanced)

For problem 13 (Single Replacement: $\ce{Cu + AgNO_{3} -> Cu(NO_{3})_{2} + Ag}$):
# Explanation:
## Step1: Count atoms
Left: Cu = 1, Ag = 1, N = 1, O = 3; Right: Cu = 1, Ag = 1, N = 2, O = 6.
## Step2: Balance N and O
Multiply $\ce{AgNO_{3}}$ by 2: $\ce{Cu + 2AgNO_{3} -> Cu(NO_{3})_{2} + Ag}$. Now N:2, O:6 on left. Right N:2, O:6. Ag:2 on left, 1 on right.
## Step3: Balance Ag
Multiply Ag by 2: $\ce{Cu + 2AgNO_{3} -> Cu(NO_{3})_{2} + 2Ag}$. Now Cu:1, Ag:2, N:2, O:6 on both sides.
# Answer:
$\ce{Cu + 2AgNO_{3} -> Cu(NO_{3})_{2} + 2Ag}$

For problem 14 (Single Replacement: $\ce{K + H_{2}O -> KOH + H_{2}}$):
# Explanation:
## Step1: Count atoms
Left: K = 1, H = 2, O = 1; Right: K = 1, H = 1 + 2 = 3, O = 1.
## Step2: Balance H
Multiply $\ce{H_{2}O}$ by 2 and $\ce{KOH}$ by 2: $\ce{K + 2H_{2}O -> 2KOH + H_{2}}$. Now H:4 on left, 2 + 2 = 4 on right. O:2 on left, 2 on right. K:1 on left, 2 on right.
## Step3: Balance K
Multiply K by 2: $\ce{2K + 2H_{2}O -> 2KOH + H_{2}}$. Now K:2, H:4, O:2 on both sides.
# Answer:
$\ce{2K + 2H_{2}O -> 2KOH + H_{2}}$

For problem 15 (Single Replacement: $\ce{Al + CuCl_{2} -> AlCl_{3} + Cu}$):
# Explanation:
## Step1: Count atoms
Left: Al = 1, Cu = 1, Cl = 2; Right: Al = 1, Cu = 1, Cl = 3.
## Step2: Balance Cl
Multiply $\ce{CuCl_{2}}$ by 3 and $\ce{AlCl_{3}}$ by 2: $\ce{Al + 3CuCl_{2} -> 2AlCl_{3} + Cu}$. Now Cl:6 on both sides. Al:1 on left, 2 on right. Cu:3 on left, 1 on right.
## Step3: Balance Al and Cu
Multiply Al by 2 and Cu by 3: $\ce{2Al + 3CuCl_{2} -> 2AlCl_{3} + 3Cu}$. Now Al:2, Cu:3, Cl:6 on both sides.
# Answer:
$\ce{2Al + 3CuCl_{2} -> 2AlCl_{3} + 3Cu}$

For problem 16 (Double Replacement: $\ce{BaCl_{2} + Na_{2}SO_{4} -> NaCl + BaSO_{4}}$):
# Explanation:
## Step1: Count atoms
Left: Ba = 1, Cl = 2, Na = 2, S = 1, O = 4; Right: Ba = 1, Cl = 1, Na = 1, S = 1, O = 4.
## Step2: Balance Na and Cl
Multiply $\ce{NaCl}$ by 2: $\ce{BaCl_{2} + Na_{2}SO_{4} -> 2NaCl + BaSO_{4}}$. Now Ba:1, Cl:2, Na:2, S:1, O:4 on both sides.
# Answer:
$\ce{BaCl_{2} + Na_{2}SO_{4} -> 2NaCl + BaSO_{4}}$ (already balanced after multiplying NaCl by 2)

For problem 17 (Double Replacement: $\ce{ZnCl_{2} + (NH_{4})_{2}S -> NH_{4}Cl + ZnS}$):
# Explanation:
## Step1: Count atoms
Left: Zn = 1, Cl = 2, NH₄ = 2, S = 1; Right: Zn = 1, Cl = 1, NH₄ = 1, S = 1.
## Step2: Balance NH₄ and Cl
Multiply $\ce{NH_{4}Cl}$ by 2: $\ce{ZnCl_{2} + (NH_{4})_{2}S -> 2NH_{4}Cl + ZnS}$. Now Zn:1, Cl:2, NH₄:2, S:1 on both sides.
# Answer:
$\ce{ZnCl_{2} + (NH_{4})_{2}S -> 2NH_{4}Cl + ZnS}$ (already balanced)

For problem 18 (Double Replacement: $\ce{NaOH + HCl -> NaCl + H_{2}O}$):
# Explanation:
## Step1: Count atoms
Left: Na = 1, O = 1, H = 1 + 1 = 2, Cl = 1; Right: Na = 1, Cl = 1, H = 2, O = 1. Already balanced.
# Answer:
$\ce{NaOH + HCl -> NaCl + H_{2}O}$ (already balanced)

For problem 19 (Double Replacement: $\ce{FeS + HCl -> FeCl_{2} + H_{2}S}$):
# Explanation:
## Step1: Count atoms
Left: Fe = 1, S = 1, H = 1, Cl = 1; Right: Fe = 1, S = 1, H = 2, Cl = 2.
## Step2: Balance H and Cl
Multiply HCl by 2: $\ce{FeS + 2HCl -> FeCl_{2} + H_{2}S}$. Now Fe:1, S:1, H:2, Cl:2 on both sides.
# Answer:
$\ce{FeS + 2HCl -> FeCl_{2} + H_{2}S}$

For problem 20 (Double Replacement: $\ce{AlCl_{3} + NaOH -> }$ (products? Probably $\ce{Al(OH)_{3} + NaCl}$). Let's assume the

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QUESTION IMAGE

Question

Explanation:

Step1: Count atoms on each side

Step2: Balance O atoms

Step3: Balance Ba atoms

Step1: Count atoms

Step2: Balance S

Step1: Count atoms

Step2: Balance O

Step3: Balance Li

Answer: