8) \underline{\quad\quad} c₄h₈ + \underline{\quad\quad} o₂ → \underline{\quad\quad} co₂ + \underline{\quad\quad} h₂o\
\
9) \underline{\quad\quad} ga(oh)₃ + \underline{\quad\quad} kf → \underline{\quad\quad} koh + \underline{\quad\quad} gaf₃\
\
10) \underline{\quad\quad} v + \underline{\quad\quad} znbr₂ → \underline{\quad\quad} vbr₃ + \underline{\quad\quad} zn\
\
11) \underline{\quad\quad} as₂o₅ + \underline{\quad\quad} h₂o → \underline{\quad\quad} h₃aso₄\
\
12) \underline{\quad\quad} nh₃ + \underline{\quad\quad} o₂ → \underline{\quad\quad} no + \underline{\quad\quad} h₂o\
\
13) \underline{\quad\quad} c₃h₈ + \underline{\quad\quad} o₂ → \underline{\quad\quad} co₂ + \underline{\quad\quad} h₂o\
\
14) \underline{\quad\quad} naclo₃ → \underline{\quad\quad} nacl + \underline{\quad\quad} o₂\
\
15) \underline{\quad\quad} ca + \underline{\quad\quad} o₂ → \underline{\quad\quad} cao

Question

Accepted Answer

### Problem 8:
# Explanation:
## Step1: Balance Carbon atoms
There are 4 C in $ C_4H_8 $, so we put 4 in front of $ CO_2 $.
$ 1\ C_4H_8 + \_\_ O_2
ightarrow 4\ CO_2 + \_\_ H_2O $

## Step2: Balance Hydrogen atoms
There are 8 H in $ C_4H_8 $, so we put 4 in front of $ H_2O $ (since each $ H_2O $ has 2 H).
$ 1\ C_4H_8 + \_\_ O_2
ightarrow 4\ CO_2 + 4\ H_2O $

## Step3: Balance Oxygen atoms
On the right, we have $ 4\times2 + 4\times1 = 12 $ O. So we need 6 $ O_2 $ (since each $ O_2 $ has 2 O).
$ 1\ C_4H_8 + 6\ O_2
ightarrow 4\ CO_2 + 4\ H_2O $

# Answer:
$ 1\ C_4H_8 + 6\ O_2
ightarrow 4\ CO_2 + 4\ H_2O $

### Problem 9:
# Explanation:
## Step1: Balance Fluorine atoms
There are 3 F in $ GaF_3 $, so we put 3 in front of $ KF $.
$ \_\_ Ga(OH)_3 + 3\ KF
ightarrow \_\_ KOH + 1\ GaF_3 $

## Step2: Balance Potassium atoms
There are 3 K in $ 3\ KF $, so we put 3 in front of $ KOH $.
$ \_\_ Ga(OH)_3 + 3\ KF
ightarrow 3\ KOH + 1\ GaF_3 $

## Step3: Balance Gallium and Oxygen/Hydrogen atoms
There is 1 Ga in $ Ga(OH)_3 $ and 1 in $ GaF_3 $, so we put 1 in front of $ Ga(OH)_3 $. The O and H atoms are already balanced (3 OH in $ Ga(OH)_3 $ and 3 OH in $ 3\ KOH $).
$ 1\ Ga(OH)_3 + 3\ KF
ightarrow 3\ KOH + 1\ GaF_3 $

# Answer:
$ 1\ Ga(OH)_3 + 3\ KF
ightarrow 3\ KOH + 1\ GaF_3 $

### Problem 10:
# Explanation:
## Step1: Balance Bromine atoms
There are 3 Br in $ VBr_3 $ and 2 Br in $ ZnBr_2 $. The least common multiple of 3 and 2 is 6. So we put 2 in front of $ VBr_3 $ and 3 in front of $ ZnBr_2 $.
$ \_\_ V + 3\ ZnBr_2
ightarrow 2\ VBr_3 + \_\_ Zn $

## Step2: Balance Vanadium atoms
There are 2 V in $ 2\ VBr_3 $, so we put 2 in front of $ V $.
$ 2\ V + 3\ ZnBr_2
ightarrow 2\ VBr_3 + \_\_ Zn $

## Step3: Balance Zinc atoms
There are 3 Zn in $ 3\ ZnBr_2 $, so we put 3 in front of $ Zn $.
$ 2\ V + 3\ ZnBr_2
ightarrow 2\ VBr_3 + 3\ Zn $

# Answer:
$ 2\ V + 3\ ZnBr_2
ightarrow 2\ VBr_3 + 3\ Zn $

### Problem 11:
# Explanation:
## Step1: Balance Arsenic atoms
There are 2 As in $ As_2O_5 $, so we put 2 in front of $ H_3AsO_4 $.
$ 1\ As_2O_5 + \_\_ H_2O
ightarrow 2\ H_3AsO_4 $

## Step2: Balance Hydrogen and Oxygen atoms
On the right, we have $ 2\times3 = 6 $ H, so we put 3 in front of $ H_2O $ (since each $ H_2O $ has 2 H). The O atoms: $ 5 + 3 = 8 $ on the left, and $ 2\times4 = 8 $ on the right, so they are balanced.
$ 1\ As_2O_5 + 3\ H_2O
ightarrow 2\ H_3AsO_4 $

# Answer:
$ 1\ As_2O_5 + 3\ H_2O
ightarrow 2\ H_3AsO_4 $

### Problem 12:
# Explanation:
## Step1: Balance Nitrogen atoms
There are 1 N in $ NH_3 $ and 1 N in $ NO $, so we put 4 in front of $ NH_3 $ and 4 in front of $ NO $ (to balance with O later).
$ 4\ NH_3 + \_\_ O_2
ightarrow 4\ NO + \_\_ H_2O $

## Step2: Balance Hydrogen atoms
There are $ 4\times3 = 12 $ H in $ 4\ NH_3 $, so we put 6 in front of $ H_2O $ (since each $ H_2O $ has 2 H).
$ 4\ NH_3 + \_\_ O_2
ightarrow 4\ NO + 6\ H_2O $

## Step3: Balance Oxygen atoms
On the right, we have $ 4\times1 + 6\times1 = 10 $ O. So we need 5 $ O_2 $ (since each $ O_2 $ has 2 O).
$ 4\ NH_3 + 5\ O_2
ightarrow 4\ NO + 6\ H_2O $

# Answer:
$ 4\ NH_3 + 5\ O_2
ightarrow 4\ NO + 6\ H_2O $

### Problem 13:
# Explanation:
## Step1: Balance Carbon atoms
There are 3 C in $ C_3H_8 $, so we put 3 in front of $ CO_2 $.
$ 1\ C_3H_8 + \_\_ O_2
ightarrow 3\ CO_2 + \_\_ H_2O $

## Step2: Balance Hydrogen atoms
There are 8 H in $ C_3H_8 $, so we put 4 in front of $ H_2O $ (since each $ H_2O $ has 2 H).
$ 1\ C_3H_8 + \_\_ O_2
ightarrow 3\ CO_2 + 4\ H_2O $

## Step3: Balance Oxygen atoms
On the right, we have $ 3\times2 + 4\times1 = 10 $ O. So we need 5 $ O_2 $ (since each $ O_2 $ has 2 O).
$ 1\ C_3H_8 + 5\ O_2
ightarrow 3\ CO_2 + 4\ H_2O $

# Answer:
$ 1\ C_3H_8 + 5\ O_2
ightarrow 3\ CO_2 + 4\ H_2O $

### Problem 14:
# Explanation:
## Step1: Balance Chlorine atoms
There is 1 Cl in $ NaClO_3 $ and 1 in $ NaCl $, so they are balanced for now.

## Step2: Balance Oxygen atoms
There are 3 O in $ NaClO_3 $, so we put $ \frac{3}{2} $ in front of $ O_2 $, but we want whole numbers. Multiply all coefficients by 2.
$ 2\ NaClO_3
ightarrow 2\ NaCl + 3\ O_2 $

## Step3: Check balance
Cl: 2 on left, 2 on right. Na: 2 on left, 2 on right. O: $ 2\times3 = 6 $ on left, $ 3\times2 = 6 $ on right. Balanced.
$ 2\ NaClO_3
ightarrow 2\ NaCl + 3\ O_2 $

# Answer:
$ 2\ NaClO_3
ightarrow 2\ NaCl + 3\ O_2 $

### Problem 15:
# Explanation:
## Step1: Balance Oxygen atoms
There are 2 O in $ O_2 $ and 1 in $ CaO $, so we put 2 in front of $ CaO $.
$ \_\_ Ca + \_\_ O_2
ightarrow 2\ CaO $

## Step2: Balance Calcium atoms
There are 2 Ca in $ 2\ CaO $, so we put 2 in front of $ Ca $.
$ 2\ Ca + \_\_ O_2
ightarrow 2\ CaO $

## Step3: Balance Oxygen atoms
There are 2 O in $ 2\ CaO $, so we put 1 in front of $ O_2 $ (since each $ O_2 $ has 2 O, 1 $ O_2 $ gives 2 O).
$ 2\ Ca + 1\ O_2
ightarrow 2\ CaO $

# Answer:
$ 2\ Ca + 1\ O_2
ightarrow 2\ CaO $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 8:

Step1: Balance Carbon atoms

Step2: Balance Hydrogen atoms

Step3: Balance Oxygen atoms

Step1: Balance Fluorine atoms

Step2: Balance Potassium atoms

Step3: Balance Gallium and Oxygen/Hydrogen atoms

Step1: Balance Bromine atoms

Step2: Balance Vanadium atoms

Step3: Balance Zinc atoms

Answer:

Problem 9: