8.\underline{\quad\quad} al + \underline{\quad\quad} fe₂o₃ → \underline{\quad\quad} al₂o₃ + \underline{\quad\quad} fe\
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9.\underline{\quad\quad} na + \underline{\quad\quad} i₂ → \underline{\quad\quad} nai\
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10.\underline{\quad\quad} mg + \underline{\quad\quad} o₂ → \underline{\quad\quad} mgo\
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11.\underline{\quad\quad} na₂co₃ + \underline{\quad\quad} cacl₂ →\underline{\quad\quad} caco₃ + \underline{\quad\quad} nacl\
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12.\underline{\quad\quad} agno₃ + \underline{\quad\quad} zncl₂ → \underline{\quad\quad} agcl + \underline{\quad\quad} zn(no₃)₂\
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13.\underline{\quad\quad} feso₄ + \underline{\quad\quad} (nh₄)₂s → \underline{\quad\quad} fes + \underline{\quad\quad} (nh₄)₂so₄\
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14. \underline{\quad\quad} mg + \underline{\quad\quad} h₃po₄ → \underline{\quad\quad} mg₃(po₄)₂ + \underline{\quad\quad} h₂\
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15. \underline{\quad\quad} cl₂ + \underline{\quad\quad} ali₃ → \underline{\quad\quad} alcl₃ + \underline{\quad\quad} i₂

Question

Accepted Answer

### Problem 8:
# Explanation:
## Step1: Balance Al atoms.
On the left, we have 1 Al, on the right, 2 Al in $Al_2O_3$. So we put a coefficient of 2 in front of Al: $2Al + Fe_2O_3
ightarrow Al_2O_3 + Fe$
## Step2: Balance Fe atoms.
On the left, 2 Fe in $Fe_2O_3$, on the right, 1 Fe. So we put a coefficient of 2 in front of Fe: $2Al + Fe_2O_3
ightarrow Al_2O_3 + 2Fe$
## Step3: Check O atoms.
Left: 3 O (from $Fe_2O_3$), Right: 3 O (from $Al_2O_3$). Balanced.
# Answer:
2, 1, 1, 2

### Problem 9:
# Explanation:
## Step1: Balance I atoms.
On the left, 2 I in $I_2$, on the right, 1 I in NaI. So we put a coefficient of 2 in front of NaI: $Na + I_2
ightarrow 2NaI$
## Step2: Balance Na atoms.
On the left, 1 Na, on the right, 2 Na (from 2NaI). So we put a coefficient of 2 in front of Na: $2Na + I_2
ightarrow 2NaI$
# Answer:
2, 1, 2

### Problem 10:
# Explanation:
## Step1: Balance O atoms.
On the left, 2 O in $O_2$, on the right, 1 O in MgO. So we put a coefficient of 2 in front of MgO: $Mg + O_2
ightarrow 2MgO$
## Step2: Balance Mg atoms.
On the left, 1 Mg, on the right, 2 Mg (from 2MgO). So we put a coefficient of 2 in front of Mg: $2Mg + O_2
ightarrow 2MgO$
# Answer:
2, 1, 2

### Problem 11:
# Explanation:
## Step1: Balance Ca atoms.
On the left, 1 Ca in $CaCl_2$, on the right, 1 Ca in $CaCO_3$. Balanced for Ca.
## Step2: Balance Na atoms.
On the left, 2 Na in $Na_2CO_3$, on the right, 1 Na in NaCl. So we put a coefficient of 2 in front of NaCl: $Na_2CO_3 + CaCl_2
ightarrow CaCO_3 + 2NaCl$
## Step3: Check Cl atoms.
Left: 2 Cl (from $CaCl_2$), Right: 2 Cl (from 2NaCl). Balanced.
## Step4: Check C and O atoms.
Left: 1 C, 3 O (from $Na_2CO_3$), Right: 1 C, 3 O (from $CaCO_3$). Balanced.
# Answer:
1, 1, 1, 2

### Problem 12:
# Explanation:
## Step1: Balance NO₃⁻ ions.
On the left, 1 $NO_3^-$ in $AgNO_3$, on the right, 2 $NO_3^-$ in $Zn(NO_3)_2$. So we put a coefficient of 2 in front of $AgNO_3$: $2AgNO_3 + ZnCl_2
ightarrow AgCl + Zn(NO_3)_2$
## Step2: Balance Ag atoms.
On the left, 2 Ag (from 2$AgNO_3$), on the right, 1 Ag in AgCl. So we put a coefficient of 2 in front of AgCl: $2AgNO_3 + ZnCl_2
ightarrow 2AgCl + Zn(NO_3)_2$
## Step3: Check Cl atoms.
Left: 2 Cl (from $ZnCl_2$), Right: 2 Cl (from 2AgCl). Balanced.
## Step4: Check Zn atoms.
Left: 1 Zn, Right: 1 Zn. Balanced.
# Answer:
2, 1, 2, 1

### Problem 13:
# Explanation:
## Step1: Check all atoms.
Fe: Left 1 (from $FeSO_4$), Right 1 (from FeS). Balanced.
S: Left 1 (from $FeSO_4$) and 1 (from $(NH_4)_2S$), Right 1 (from FeS) and 1 (from $(NH_4)_2SO_4$). Balanced.
N: Left 2 (from $(NH_4)_2S$), Right 2 (from $(NH_4)_2SO_4$). Balanced.
H: Left 8 (from $(NH_4)_2S$), Right 8 (from $(NH_4)_2SO_4$). Balanced.
O: Left 4 (from $FeSO_4$), Right 4 (from $(NH_4)_2SO_4$). Balanced.
So the coefficients are all 1.
# Answer:
1, 1, 1, 1

### Problem 14:
# Explanation:
## Step1: Balance Mg atoms.
On the left, 1 Mg, on the right, 3 Mg in $Mg_3(PO_4)_2$. So we put a coefficient of 3 in front of Mg: $3Mg + H_3PO_4
ightarrow Mg_3(PO_4)_2 + H_2$
## Step2: Balance PO₄³⁻ ions.
On the left, 1 $PO_4^{3-}$ in $H_3PO_4$, on the right, 2 $PO_4^{3-}$ in $Mg_3(PO_4)_2$. So we put a coefficient of 2 in front of $H_3PO_4$: $3Mg + 2H_3PO_4
ightarrow Mg_3(PO_4)_2 + H_2$
## Step3: Balance H atoms.
On the left, 6 H (from 2$H_3PO_4$), on the right, 2 H in $H_2$. So we put a coefficient of 3 in front of $H_2$: $3Mg + 2H_3PO_4
ightarrow Mg_3(PO_4)_2 + 3H_2$
# Answer:
3, 2, 1, 3

### Problem 15:
# Explanation:
## Step1: Balance Cl atoms.
On the left, 2 Cl in $Cl_2$, on the right, 3 Cl in $AlCl_3$. The least common multiple of 2 and 3 is 6. So we put a coefficient of 3 in front of $Cl_2$ and 2 in front of $AlCl_3$: $3Cl_2 + AlI_3
ightarrow 2AlCl_3 + I_2$
## Step2: Balance Al atoms.
On the left, 1 Al in $AlI_3$, on the right, 2 Al in 2$AlCl_3$. So we put a coefficient of 2 in front of $AlI_3$: $3Cl_2 + 2AlI_3
ightarrow 2AlCl_3 + I_2$
## Step3: Balance I atoms.
On the left, 6 I (from 2$AlI_3$), on the right, 2 I in $I_2$. So we put a coefficient of 3 in front of $I_2$: $3Cl_2 + 2AlI_3
ightarrow 2AlCl_3 + 3I_2$
# Answer:
3, 2, 2, 3

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QUESTION IMAGE

Question

Explanation:

Problem 8:

Step1: Balance Al atoms.

Step2: Balance Fe atoms.

Step3: Check O atoms.

Step1: Balance I atoms.

Step2: Balance Na atoms.

Step1: Balance O atoms.

Step2: Balance Mg atoms.

Answer:

Problem 9: