9. \underline{\quad\quad} fecl₃ + \underline{\quad\quad} naoh \
ightarrow \underline{\quad\quad} fe(oh)₃ + \underline{\quad\quad}nacl\
\
10. \underline{\quad\quad} p + \underline{\quad\quad}o₂ \
ightarrow \underline{\quad\quad}p₂o₅\
\
11. \underline{\quad\quad} na + \underline{\quad\quad} h₂o \
ightarrow \underline{\quad\quad} naoh + \underline{\quad\quad}h₂\
\
12. \underline{\quad\quad} s₈ + \underline{\quad\quad}o₂ \
ightarrow \underline{\quad\quad} so₃\
\
13. \underline{\quad\quad} co₂ + \underline{\quad\quad} h₂o \
ightarrow \underline{\quad\quad} c₆h₁₂o₆ + \underline{\quad\quad}o₂\
\
14. \underline{\quad\quad} k + \underline{\quad\quad} mgbr \
ightarrow \underline{\quad\quad} kbr + \underline{\quad\quad} mg\
\
15. \underline{\quad\quad} hcl + \underline{\quad\quad} caco₃ \
ightarrow \underline{\quad\quad} cacl₂ + \underline{\quad\quad}h₂o + \underline{\quad\quad} co₂\
\
16. \underline{\quad\quad} hno₃ + \underline{\quad\quad} nahco₃ \
ightarrow \underline{\quad\quad} nano₃ + \underline{\quad\quad} h₂o + \underline{\quad\quad} co₂\
\
17. \underline{\quad\quad} h₂o + \underline{\quad\quad} o₂ \
ightarrow \underline{\quad\quad} h₂o₂\
\
18. \underline{\quad\quad} nabr + \underline{\quad\quad} caf₂ \
ightarrow \underline{\quad\quad} naf + \underline{\quad\quad} cabr₂

Question

Accepted Answer

### Problem 9
# Explanation:
## Step1: Balance Fe
There is 1 Fe in $ \ce{FeCl3} $ and 1 Fe in $ \ce{Fe(OH)3} $, so Fe is balanced.
## Step2: Balance Cl
There are 3 Cl in $ \ce{FeCl3} $ and 1 Cl in $ \ce{NaCl} $. So we need 3 $ \ce{NaCl} $, which gives 3 Na.
## Step3: Balance Na
There are 3 Na in $ \ce{NaCl} $, so we need 3 $ \ce{NaOH} $, which gives 3 OH.
## Step4: Balance OH
There are 3 OH in $ \ce{NaOH} $ and 3 OH in $ \ce{Fe(OH)3} $, so OH is balanced.
So the balanced equation is $ \ce{1 FeCl3 + 3 NaOH -> 1 Fe(OH)3 + 3 NaCl} $
# Answer:
1, 3, 1, 3

### Problem 10
# Explanation:
## Step1: Balance P
There are 2 P in $ \ce{P2O5} $, so we need 2 P on the left, so coefficient of P is 2.
## Step2: Balance O
There are 5 O in $ \ce{P2O5} $ and 2 O in $ \ce{O2} $. The least common multiple of 5 and 2 is 10. So we need 5 $ \ce{O2} $ (10 O) and 2 $ \ce{P2O5} $ (10 O). But wait, if we have 2 $ \ce{P2O5} $, then P would be 4. So let's adjust. Let's balance O first. The number of O in $ \ce{P2O5} $ is 5, so coefficient of $ \ce{O2} $ is 5/2? No, we need whole numbers. So multiply all coefficients by 2. So P becomes 4, $ \ce{O2} $ becomes 5, and $ \ce{P2O5} $ becomes 2. Wait, no, original: $ \ce{P + O2 -> P2O5} $. Let's do it properly. Let coefficient of P be x, $ \ce{O2} $ be y, $ \ce{P2O5} $ be z. So x P, 2y O, 2z P and 5z O. So x = 2z (P balance), 2y = 5z (O balance). Let z = 2, then x = 4, y = 5. So the balanced equation is $ \ce{4 P + 5 O2 -> 2 P2O5} $
# Answer:
4, 5, 2

### Problem 11
# Explanation:
## Step1: Balance H
There are 2 H in $ \ce{H2} $ and 1 H in $ \ce{NaOH} $. Let's balance H first. Let coefficient of $ \ce{H2} $ be 1, then we need 2 $ \ce{NaOH} $ (2 H), which gives 2 Na.
## Step2: Balance Na
There are 2 Na in $ \ce{NaOH} $, so coefficient of Na is 2.
## Step3: Balance O
There are 2 O in $ \ce{NaOH} $ and 2 O in $ \ce{H2O} $ (since coefficient of $ \ce{H2O} $ is 2).
## Step4: Balance H (again)
There are 2 $ \ce{H2O} $ (4 H) and 2 $ \ce{NaOH} $ (2 H) and 1 $ \ce{H2} $ (2 H). Wait, 4 H from $ \ce{H2O} $, 2 H from $ \ce{NaOH} $? No, wait, the equation is $ \ce{Na + H2O -> NaOH + H2} $. Let's do it with variables. Let Na be a, $ \ce{H2O} $ be b, $ \ce{NaOH} $ be c, $ \ce{H2} $ be d. So a Na, 2b H, b O, c Na, c O, c H, 2d H. So Na: a = c. O: b = c. H: 2b = c + 2d. Since a = c = b, substitute: 2b = b + 2d => b = 2d. Let d = 1, then b = 2, c = 2, a = 2. So the balanced equation is $ \ce{2 Na + 2 H2O -> 2 NaOH + 1 H2} $
# Answer:
2, 2, 2, 1

### Problem 12
# Explanation:
## Step1: Balance S
There are 8 S in $ \ce{S8} $, so we need 8 $ \ce{SO3} $, which gives 24 O.
## Step2: Balance O
There are 24 O in $ \ce{SO3} $ and 2 O in $ \ce{O2} $, so we need 12 $ \ce{O2} $.
So the balanced equation is $ \ce{1 S8 + 12 O2 -> 8 SO3} $
# Answer:
1, 12, 8

### Problem 13
# Explanation:
## Step1: Balance C
There are 6 C in $ \ce{C6H12O6} $, so we need 6 $ \ce{CO2} $, which gives 6 C.
## Step2: Balance H
There are 12 H in $ \ce{C6H12O6} $ and 2 H in $ \ce{H2O} $, so we need 6 $ \ce{H2O} $, which gives 6 O from $ \ce{H2O} $ and 12 O from $ \ce{CO2} $ (total 18 O).
## Step3: Balance O
There are 6 O in $ \ce{C6H12O6} $ and 2 O in $ \ce{O2} $. So total O on right: 6 + 2x (where x is coefficient of $ \ce{O2} $) = 18. So 2x = 12, x = 6.
So the balanced equation is $ \ce{6 CO2 + 6 H2O -> 1 C6H12O6 + 6 O2} $
# Answer:
6, 6, 1, 6

### Problem 14 (Note: The formula should be $ \ce{MgBr2} $ instead of $ \ce{MgBr} $, assuming a typo)
# Explanation:
## Step1: Balance Br
There are 2 Br in $ \ce{MgBr2} $ and 1 Br in $ \ce{KBr} $, so we need 2 $ \ce{KBr} $, which gives 2 K.
## Step2: Balance K
There are 2 K in $ \ce{KBr} $, so we need 2 K on the left, so coefficient of K is 2.
## Step3: Balance Mg
There is 1 Mg in $ \ce{MgBr2} $ and 1 Mg in $ \ce{Mg} $, so Mg is balanced.
So the balanced equation (with $ \ce{MgBr2} $) is $ \ce{2 K + 1 MgBr2 -> 2 KBr + 1 Mg} $
# Answer:
2, 1, 2, 1 (assuming $ \ce{MgBr2} $ instead of $ \ce{MgBr} $)

### Problem 15
# Explanation:
## Step1: Balance Ca
There is 1 Ca in $ \ce{CaCO3} $ and 1 Ca in $ \ce{CaCl2} $, so Ca is balanced.
## Step2: Balance Cl
There are 2 Cl in $ \ce{CaCl2} $ and 1 Cl in $ \ce{HCl} $, so we need 2 $ \ce{HCl} $, which gives 2 H.
## Step3: Balance H
There are 2 H in $ \ce{HCl} $ and 2 H in $ \ce{H2O} $ (since coefficient of $ \ce{H2O} $ is 1? Wait, no. Let's do it properly. $ \ce{HCl + CaCO3 -> CaCl2 + H2O + CO2} $. Let coefficient of $ \ce{HCl} $ be x, $ \ce{CaCO3} $ be 1, $ \ce{CaCl2} $ be 1, $ \ce{H2O} $ be y, $ \ce{CO2} $ be 1. So x Cl, 1 Ca, 1 C, 3 O, x H, 1 Cl, 2 Cl (wait, $ \ce{CaCl2} $ has 2 Cl), so x = 2. Then H: 2 H, so y = 1 (since $ \ce{H2O} $ has 2 H). O: 3 O in $ \ce{CaCO3} $, 1 O in $ \ce{H2O} $, 2 O in $ \ce{CO2} $. Wait, 3 O (from $ \ce{CaCO3} $) = 1 O (from $ \ce{H2O} $) + 2 O (from $ \ce{CO2} $)? 1 + 2 = 3, yes. C: 1 C in $ \ce{CaCO3} $ and 1 C in $ \ce{CO2} $, balanced. So the balanced equation is $ \ce{2 HCl + 1 CaCO3 -> 1 CaCl2 + 1 H2O + 1 CO2} $
# Answer:
2, 1, 1, 1, 1

### Problem 16
# Explanation:
## Step1: Balance Na
There is 1 Na in $ \ce{NaHCO3} $ and 1 Na in $ \ce{NaNO3} $, so Na is balanced.
## Step2: Balance N
There is 1 N in $ \ce{HNO3} $ and 1 N in $ \ce{NaNO3} $, so N is balanced.
## Step3: Balance C
There is 1 C in $ \ce{NaHCO3} $ and 1 C in $ \ce{CO2} $, so C is balanced.
## Step4: Balance H
There is 1 H in $ \ce{HNO3} $ and 1 H in $ \ce{NaHCO3} $, total 2 H, which is balanced by 1 $ \ce{H2O} $ (2 H? Wait, no. $ \ce{HNO3 + NaHCO3 -> NaNO3 + H2O + CO2} $. Let's check H: $ \ce{HNO3} $ has 1 H, $ \ce{NaHCO3} $ has 1 H, so total 2 H, which is in $ \ce{H2O} $ (2 H). O: $ \ce{HNO3} $ has 3 O, $ \ce{NaHCO3} $ has 3 O, total 6 O. $ \ce{NaNO3} $ has 3 O, $ \ce{H2O} $ has 1 O, $ \ce{CO2} $ has 2 O, total 6 O. So balanced. So coefficients are 1, 1, 1, 1, 1.
# Answer:
1, 1, 1, 1, 1

### Problem 17
# Explanation:
## Step1: Balance O
There are 2 O in $ \ce{O2} $ and 2 O in $ \ce{H2O2} $, but $ \ce{H2O} $ has 1 O. Wait, the equation is $ \ce{H2O + O2 -> H2O2} $. Let coefficient of $ \ce{H2O} $ be x, $ \ce{O2} $ be y, $ \ce{H2O2} $ be z. So 2x H, x + 2y O, 2z H, 2z O. So 2x = 2z (H balance) => x = z. x + 2y = 2z (O balance) => x + 2y = 2x => 2y = x. Let x = 2, then z = 2, y = 1. So the balanced equation is $ \ce{2 H2O + 1 O2 -> 2 H2O2} $
# Answer:
2, 1, 2

### Problem 18
# Explanation:
## Step1: Balance Br
There are 2 Br in $ \ce{CaBr2} $ and 1 Br in $ \ce{NaBr} $, so we need 2 $ \ce{NaBr} $, which gives 2 Na.
## Step2: Balance Na
There are 2 Na in $ \ce{NaBr} $ and 1 Na in $ \ce{NaF} $, so we need 2 $ \ce{NaF} $, which gives 2 F.
## Step3: Balance F
There are 2 F in $ \ce{NaF} $ and 2 F in $ \ce{CaF2} $, so coefficient of $ \ce{CaF2} $ is 1.
## Step4: Balance Ca
There is 1 Ca in $ \ce{CaF2} $ and 1 Ca in $ \ce{CaBr2} $, so Ca is balanced.
So the balanced equation is $ \ce{2 NaBr + 1 CaF2 -> 2 NaF + 1 CaBr2} $
# Answer:
2, 1, 2, 1

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 9

Step1: Balance Fe

Step2: Balance Cl

Step3: Balance Na

Step4: Balance OH

Step1: Balance P

Step2: Balance O

Step1: Balance H

Step2: Balance Na

Step3: Balance O

Step4: Balance H (again)

Answer:

Problem 10