8) ______ $\ce{c_{4}h_{8}}$ + ______ $\ce{o_{2}}$ $\
ightarrow$ ______ $\ce{co_{2}}$ + ______ $\ce{h_{2}o}$

9) ______ $\ce{ga(oh)_{3}}$ + ______ $\ce{kf}$ $\
ightarrow$ ______ $\ce{koh}$ + ______ $\ce{gaf_{3}}$

10) ______ $\ce{v}$ + ______ $\ce{znbr_{2}}$ $\
ightarrow$ ______ $\ce{vbr_{3}}$ + ______ $\ce{zn}$

11) ______ $\ce{as_{2}o_{5}}$ + ______ $\ce{h_{2}o}$ $\
ightarrow$ ______ $\ce{h_{3}aso_{4}}$

12) ______ $\ce{nh_{3}}$ + ______ $\ce{o_{2}}$ $\
ightarrow$ ______ $\ce{no}$ + ______ $\ce{h_{2}o}$

13) ______ $\ce{c_{3}h_{8}}$ + ______ $\ce{o_{2}}$ $\
ightarrow$ ______ $\ce{co_{2}}$ + ______ $\ce{h_{2}o}$

14) ______ $\ce{naclo_{3}}$ $\
ightarrow$ ______ $\ce{nacl}$ + ______ $\ce{o_{2}}$

15) ______ $\ce{ca}$ + ______ $\ce{o_{2}}$ $\
ightarrow$ ______ $\ce{cao}$

Question

8) ______ $\ce{c_{4}h_{8}}$ + ______ $\ce{o_{2}}$ $\
ightarrow$ ______ $\ce{co_{2}}$ + ______ $\ce{h_{2}o}$

9) ______ $\ce{ga(oh)_{3}}$ + ______ $\ce{kf}$ $\
ightarrow$ ______ $\ce{koh}$ + ______ $\ce{gaf_{3}}$

10) ______ $\ce{v}$ + ______ $\ce{znbr_{2}}$ $\
ightarrow$ ______ $\ce{vbr_{3}}$ + ______ $\ce{zn}$

11) ______ $\ce{as_{2}o_{5}}$ + ______ $\ce{h_{2}o}$ $\
ightarrow$ ______ $\ce{h_{3}aso_{4}}$

12) ______ $\ce{nh_{3}}$ + ______ $\ce{o_{2}}$ $\
ightarrow$ ______ $\ce{no}$ + ______ $\ce{h_{2}o}$

13) ______ $\ce{c_{3}h_{8}}$ + ______ $\ce{o_{2}}$ $\
ightarrow$ ______ $\ce{co_{2}}$ + ______ $\ce{h_{2}o}$

14) ______ $\ce{naclo_{3}}$ $\
ightarrow$ ______ $\ce{nacl}$ + ______ $\ce{o_{2}}$

15) ______ $\ce{ca}$ + ______ $\ce{o_{2}}$ $\
ightarrow$ ______ $\ce{cao}$

Accepted Answer

### Problem 8: Balancing $ \boldsymbol{\ce{C4H8 + O2 -> CO2 + H2O}} $
# Explanation:
## Step 1: Balance Carbon (C)
There are 4 C in $ \ce{C4H8} $, so we need 4 $ \ce{CO2} $.
$ \ce{C4H8 + O2 -> 4CO2 + H2O} $

## Step 2: Balance Hydrogen (H)
There are 8 H in $ \ce{C4H8} $, so we need 4 $ \ce{H2O} $ (since each $ \ce{H2O} $ has 2 H).
$ \ce{C4H8 + O2 -> 4CO2 + 4H2O} $

## Step 3: Balance Oxygen (O)
On the right, we have $ 4 \times 2 + 4 \times 1 = 12 $ O. So we need $ \frac{12}{2} = 6 $ $ \ce{O2} $.
$ \ce{C4H8 + 6O2 -> 4CO2 + 4H2O} $

# Answer:
$ 1 \ce{C4H8} + 6 \ce{O2} -> 4 \ce{CO2} + 4 \ce{H2O} $ (Coefficients: 1, 6, 4, 4)

### Problem 9: Balancing $ \boldsymbol{\ce{Ga(OH)3 + KF -> KOH + GaF3}} $
# Explanation:
## Step 1: Balance Fluorine (F)
There are 3 F in $ \ce{GaF3} $, so we need 3 $ \ce{KF} $.
$ \ce{Ga(OH)3 + 3KF -> KOH + GaF3} $

## Step 2: Balance Potassium (K)
There are 3 K in $ 3\ce{KF} $, so we need 3 $ \ce{KOH} $.
$ \ce{Ga(OH)3 + 3KF -> 3KOH + GaF3} $

## Step 3: Check other elements (Ga, O, H)
Ga: 1 on both sides. O: $ 3 \times 1 = 3 $ on left, $ 3 \times 1 = 3 $ on right. H: $ 3 \times 1 = 3 $ on left, $ 3 \times 1 = 3 $ on right. Balanced.

# Answer:
$ 1 \ce{Ga(OH)3} + 3 \ce{KF} -> 3 \ce{KOH} + 1 \ce{GaF3} $ (Coefficients: 1, 3, 3, 1)

### Problem 10: Balancing $ \boldsymbol{\ce{V + ZnBr2 -> VBr3 + Zn}} $
# Explanation:
## Step 1: Balance Bromine (Br)
There are 3 Br in $ \ce{VBr3} $ and 2 Br in $ \ce{ZnBr2} $. The least common multiple of 2 and 3 is 6. So we need 2 $ \ce{VBr3} $ (6 Br) and 3 $ \ce{ZnBr2} $ (6 Br).
$ \ce{V + 3ZnBr2 -> 2VBr3 + Zn} $

## Step 2: Balance Vanadium (V)
Now we have 2 V on the right, so we need 2 V on the left.
$ \ce{2V + 3ZnBr2 -> 2VBr3 + Zn} $

## Step 3: Balance Zinc (Zn)
There are 3 Zn in $ 3\ce{ZnBr2} $, so we need 3 Zn on the right.
$ \ce{2V + 3ZnBr2 -> 2VBr3 + 3Zn} $

# Answer:
$ 2 \ce{V} + 3 \ce{ZnBr2} -> 2 \ce{VBr3} + 3 \ce{Zn} $ (Coefficients: 2, 3, 2, 3)

### Problem 11: Balancing $ \boldsymbol{\ce{As2O5 + H2O -> H3AsO4}} $
# Explanation:
## Step 1: Balance Arsenic (As)
There are 2 As in $ \ce{As2O5} $, so we need 2 $ \ce{H3AsO4} $.
$ \ce{As2O5 + H2O -> 2H3AsO4} $

## Step 2: Balance Hydrogen (H)
On the right, we have $ 2 \times 3 = 6 $ H, so we need 3 $ \ce{H2O} $ (since each $ \ce{H2O} $ has 2 H).
$ \ce{As2O5 + 3H2O -> 2H3AsO4} $

## Step 3: Check Oxygen (O)
Left: $ 5 + 3 \times 1 = 8 $. Right: $ 2 \times 4 = 8 $. Balanced.

# Answer:
$ 1 \ce{As2O5} + 3 \ce{H2O} -> 2 \ce{H3AsO4} $ (Coefficients: 1, 3, 2)

### Problem 12: Balancing $ \boldsymbol{\ce{NH3 + O2 -> NO + H2O}} $
# Explanation:
## Step 1: Balance Nitrogen (N)
There is 1 N in $ \ce{NH3} $ and 1 N in $ \ce{NO} $, so we need equal coefficients for $ \ce{NH3} $ and $ \ce{NO} $. Let's start with 4 $ \ce{NH3} $ and 4 $ \ce{NO} $ (to balance H later).
$ \ce{4NH3 + O2 -> 4NO + H2O} $

## Step 2: Balance Hydrogen (H)
There are $ 4 \times 3 = 12 $ H in $ 4\ce{NH3} $, so we need 6 $ \ce{H2O} $ (since each $ \ce{H2O} $ has 2 H).
$ \ce{4NH3 + O2 -> 4NO + 6H2O} $

## Step 3: Balance Oxygen (O)
On the right, we have $ 4 \times 1 + 6 \times 1 = 10 $ O. So we need $ \frac{10}{2} = 5 $ $ \ce{O2} $.
$ \ce{4NH3 + 5O2 -> 4NO + 6H2O} $

# Answer:
$ 4 \ce{NH3} + 5 \ce{O2} -> 4 \ce{NO} + 6 \ce{H2O} $ (Coefficients: 4, 5, 4, 6)

### Problem 13: Balancing $ \boldsymbol{\ce{C3H8 + O2 -> CO2 + H2O}} $
# Explanation:
## Step 1: Balance Carbon (C)
There are 3 C in $ \ce{C3H8} $, so we need 3 $ \ce{CO2} $.
$ \ce{C3H8 + O2 -> 3CO2 + H2O} $

## Step 2: Balance Hydrogen (H)
There are 8 H in $ \ce{C3H8} $, so we need 4 $ \ce{H2O} $ (since each $ \ce{H2O} $ has 2 H).
$ \ce{C3H8 + O2 -> 3CO2 + 4H2O} $

## Step 3: Balance Oxygen (O)
On the right, we have $ 3 \times 2 + 4 \times 1 = 10 $ O. So we need $ \frac{10}{2} = 5 $ $ \ce{O2} $.
$ \ce{C3H8 + 5O2 -> 3CO2 + 4H2O} $

# Answer:
$ 1 \ce{C3H8} + 5 \ce{O2} -> 3 \ce{CO2} + 4 \ce{H2O} $ (Coefficients: 1, 5, 3, 4)

### Problem 14: Balancing $ \boldsymbol{\ce{NaClO3 -> NaCl + O2}} $
# Explanation:
## Step 1: Balance Sodium (Na) and Chlorine (Cl)
Na and Cl are already balanced (1 each on both sides).

## Step 2: Balance Oxygen (O)
There are 3 O in $ \ce{NaClO3} $, so we need $ \frac{3}{2} $ $ \ce{O2} $, but we use whole numbers. Multiply all coefficients by 2.
$ \ce{2NaClO3 -> 2NaCl + 3O2} $

# Answer:
$ 2 \ce{NaClO3} -> 2 \ce{NaCl} + 3 \ce{O2} $ (Coefficients: 2, 2, 3)

### Problem 15: Balancing $ \boldsymbol{\ce{Ca + O2 -> CaO}} $
# Explanation:
## Step 1: Balance Calcium (Ca)
There is 1 Ca in $ \ce{Ca} $ and 1 Ca in $ \ce{CaO} $, so we need equal coefficients for $ \ce{Ca} $ and $ \ce{CaO} $.

## Step 2: Balance Oxygen (O)
There are 2 O in $ \ce{O2} $, so we need 2 $ \ce{CaO} $ (2 O) and thus 2 $ \ce{Ca} $.
$ \ce{2Ca + O2 -> 2CaO} $

# Answer:
$ 2 \ce{Ca} + 1 \ce{O2} -> 2 \ce{CaO} $ (Coefficients: 2, 1, 2)

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QUESTION IMAGE

Question

Explanation:

Problem 8: Balancing \( \boldsymbol{\ce{C4H8 + O2 -> CO2 + H2O}} \)

Step 1: Balance Carbon (C)

Step 2: Balance Hydrogen (H)

Step 3: Balance Oxygen (O)

Step 1: Balance Fluorine (F)

Step 2: Balance Potassium (K)

Step 3: Check other elements (Ga, O, H)

Step 1: Balance Bromine (Br)

Step 2: Balance Vanadium (V)

Step 3: Balance Zinc (Zn)

Answer:

Problem 9: Balancing \( \boldsymbol{\ce{Ga(OH)3 + KF -> KOH + GaF3}} \)