Question

equations ws #2
name
blk
cao₂ → cao + o₂
h₂ + cl₂ → hcl
ag₂o → ag + o₂
al(oh)₃ → al₂o₃ + h₂o
mno₂ + hcl → mncl₂ + h₂o + cl₂
mgo₂ → mgo + o₂
h₂s + o₂ → s + h₂o
al + cl₂ → alcl₃
pb(no₃)₂ + licl → pbcl₂ + lino₃
al₂o₃ → al + o₂
hgo → hg + o₂
al(oh)₃ + hcl → alcl₃ + h₂o
bacl₂ + koh → ba(oh)₂ + kcl
h₂o₂ → h₂o + o₂
sb₄o₆ + c → sb + co

Answer:

To solve these chemical equations, we need to balance them by ensuring the number of each type of atom is the same on both sides of the arrow. Here are the balanced equations for each:

1. $\boldsymbol{CaO_2

ightarrow CaO + O_2}$

Step 1: Count atoms

• Left: Ca = 1, O = 2
• Right: Ca = 1, O = 1 (from CaO) + 2 (from $O_2$) = 3

Step 2: Balance O

Multiply $CaO_2$ by 2, $CaO$ by 2:
$2CaO_2
ightarrow 2CaO + O_2$
Now:

• Left: Ca = 2, O = 4
• Right: Ca = 2, O = 2 (from 2CaO) + 2 (from $O_2$) = 4

Balanced: $\boldsymbol{2CaO_2
ightarrow 2CaO + O_2}$

2. $\boldsymbol{H_2 + Cl_2

ightarrow HCl}$

Step 1: Count atoms

• Left: H = 2, Cl = 2
• Right: H = 1, Cl = 1

Step 2: Balance H and Cl

Multiply $HCl$ by 2:
$H_2 + Cl_2
ightarrow 2HCl$
Balanced: $\boldsymbol{H_2 + Cl_2
ightarrow 2HCl}$

3. $\boldsymbol{Ag_2O

ightarrow Ag + O_2}$

Step 1: Count atoms

• Left: Ag = 2, O = 1
• Right: Ag = 1, O = 2

Step 2: Balance O

Multiply $Ag_2O$ by 2: $2Ag_2O
ightarrow Ag + O_2$
Now O: Left = 2, Right = 2

Step 3: Balance Ag

Multiply $Ag$ by 4:
$2Ag_2O
ightarrow 4Ag + O_2$
Balanced: $\boldsymbol{2Ag_2O
ightarrow 4Ag + O_2}$

4. $\boldsymbol{Al(OH)_3

ightarrow Al_2O_3 + H_2O}$

Step 1: Count atoms

• Left: Al = 1, O = 3, H = 3
• Right: Al = 2, O = 3 (from $Al_2O_3$) + 1 (from $H_2O$) = 4, H = 2

Step 2: Balance Al

Multiply $Al(OH)_3$ by 2: $2Al(OH)_3
ightarrow Al_2O_3 + H_2O$
Now Al: Left = 2, Right = 2

Step 3: Balance H and O

Multiply $H_2O$ by 3:
$2Al(OH)_3
ightarrow Al_2O_3 + 3H_2O$
Check:

• Left: Al = 2, O = 6, H = 6
• Right: Al = 2, O = 3 + 3 = 6, H = 6

Balanced: $\boldsymbol{2Al(OH)_3
ightarrow Al_2O_3 + 3H_2O}$

5. $\boldsymbol{MnO_2 + HCl

ightarrow MnCl_2 + H_2O + Cl_2}$

Step 1: Count atoms

• Left: Mn = 1, O = 2, H = 1, Cl = 1
• Right: Mn = 1, O = 1, H = 2, Cl = 2 (from $MnCl_2$) + 2 (from $Cl_2$) = 4

Step 2: Balance O

Multiply $H_2O$ by 2: $MnO_2 + HCl
ightarrow MnCl_2 + 2H_2O + Cl_2$
Now O: Left = 2, Right = 2

Step 3: Balance H

Multiply $HCl$ by 4: $MnO_2 + 4HCl
ightarrow MnCl_2 + 2H_2O + Cl_2$
Now H: Left = 4, Right = 4

Step 4: Balance Cl

Left Cl = 4, Right Cl = 2 (from $MnCl_2$) + 2 (from $Cl_2$) = 4
Balanced: $\boldsymbol{MnO_2 + 4HCl
ightarrow MnCl_2 + 2H_2O + Cl_2}$

6. $\boldsymbol{MgO_2

ightarrow MgO + O_2}$

Step 1: Count atoms

• Left: Mg = 1, O = 2
• Right: Mg = 1, O = 1 (from MgO) + 2 (from $O_2$) = 3

Step 2: Balance O

Multiply $MgO_2$ by 2, $MgO$ by 2:
$2MgO_2
ightarrow 2MgO + O_2$
Check:

• Left: Mg = 2, O = 4
• Right: Mg = 2, O = 2 + 2 = 4

Balanced: $\boldsymbol{2MgO_2
ightarrow 2MgO + O_2}$

7. $\boldsymbol{H_2S + O_2

ightarrow S + H_2O}$

Step 1: Count atoms

• Left: H = 2, S = 1, O = 2
• Right: H = 2, S = 1, O = 1

Step 2: Balance O

Multiply $H_2O$ by 2: $H_2S + O_2
ightarrow S + 2H_2O$
Now O: Left = 2, Right = 2

Step 3: Balance H

Left H = 2, Right H = 4 → Multiply $H_2S$ by 2: $2H_2S + O_2
ightarrow S + 2H_2O$
Now H: Left = 4, Right = 4

Step 4: Balance S

Multiply $S$ by 2: $2H_2S + O_2
ightarrow 2S + 2H_2O$
Now S: Left = 2, Right = 2

Step 5: Balance O

Multiply $O_2$ by 2: $2H_2S + 2O_2
ightarrow 2S + 2H_2O$? No, adjust:
Correct: $2H_2S + O_2
ightarrow 2S + 2H_2O$ (O: Left = 2, Right = 2)
Wait, O: Left = 2, Right = 2 (from $H_2O$). Yes.
Balanced: $\boldsymbol{2H_2S + O_2
ightarrow 2S + 2H_2O}$

8. $\boldsymbol{Al + Cl_2

ightarrow AlCl_3}$

Step 1: Count atoms

• Left: Al = 1, Cl = 2
• Right: Al = 1, Cl = 3

Step 2: Balance Cl

Multiply $Cl_2$ by 3, $AlCl_3$ by 2: $Al + 3Cl_2
ightarrow 2AlCl_3$

Step 3: Balance Al

Multiply $Al$ by 2:
$2Al + 3Cl_2
ightarrow 2AlCl_3$
Balanced: $\boldsymbol{2Al + 3Cl_2
ightarrow 2AlCl_3}$

9. $\boldsymbol{Pb(NO_3)_2 + LiCl

ightarrow PbCl_2 + LiNO_3}$

Step 1: Count atoms

• Left: Pb = 1, N = 2, O = 6, Li = 1, Cl = 1
• Right: Pb = 1, N = 1, O = 3, Li = 1, Cl = 2

Step 2: Balance N and Cl

Multiply $LiCl$ by 2, $LiNO_3$ by 2:
$Pb(NO_3)_2 + 2LiCl
ightarrow PbCl_2 + 2LiNO_3$
Check:

• Left: Pb = 1, N = 2, O = 6, Li = 2, Cl = 2
• Right: Pb = 1, N = 2, O = 6, Li = 2, Cl = 2

Balanced: $\boldsymbol{Pb(NO_3)_2 + 2LiCl
ightarrow PbCl_2 + 2LiNO_3}$

10. $\boldsymbol{Al_2O_3

ightarrow Al + O_2}$

Step 1: Count atoms

• Left: Al = 2, O = 3
• Right: Al = 1, O = 2

Step 2: Balance Al

Multiply $Al$ by 4: $Al_2O_3
ightarrow 4Al + O_2$

Step 3: Balance O

Multiply $Al_2O_3$ by 2: $2Al_2O_3
ightarrow 4Al + O_2$
Now O: Left = 6, Right = 2 → Multiply $O_2$ by 3:
$2Al_2O_3
ightarrow 4Al + 3O_2$
Check:

• Left: Al = 4, O = 6
• Right: Al = 4, O = 6

Balanced: $\boldsymbol{2Al_2O_3
ightarrow 4Al + 3O_2}$

11. $\boldsymbol{HgO

ightarrow Hg + O_2}$

Step 1: Count atoms

• Left: Hg = 1, O = 1
• Right: Hg = 1, O = 2

Step 2: Balance O

Multiply $HgO$ by 2: $2HgO
ightarrow Hg + O_2$

Step 3: Balance Hg

Multiply $Hg$ by 2:
$2HgO
ightarrow 2Hg + O_2$
Balanced: $\boldsymbol{2HgO
ightarrow 2Hg + O_2}$

12. $\boldsymbol{Al(OH)_3 + HCl

ightarrow AlCl_3 + H_2O}$

Step 1: Count atoms

• Left: Al = 1, O = 3, H = 3 + 1 = 4, Cl = 1
• Right: Al = 1, O = 1, H = 2, Cl = 3

Step 2: Balance Cl

Multiply $HCl$ by 3: $Al(OH)_3 + 3HCl
ightarrow AlCl_3 + H_2O$

Step 3: Balance H and O

Multiply $H_2O$ by 3:
$Al(OH)_3 + 3HCl
ightarrow AlCl_3 + 3H_2O$
Check:

• Left: Al = 1, O = 3, H = 3 + 3 = 6, Cl = 3
• Right: Al = 1, O = 3, H = 6, Cl = 3

Balanced: $\boldsymbol{Al(OH)_3 + 3HCl
ightarrow AlCl_3 + 3H_2O}$

13. $\boldsymbol{BaCl_2 + KOH

ightarrow Ba(OH)_2 + KCl}$

Step 1: Count atoms

• Left: Ba = 1, Cl = 2, K = 1, O = 1, H = 1
• Right: Ba = 1, Cl = 1, K = 1, O = 2, H = 2

Step 2: Balance OH⁻ (O and H)

Multiply $KOH$ by 2: $BaCl_2 + 2KOH
ightarrow Ba(OH)_2 + KCl$

Step 3: Balance K and Cl

Multiply $KCl$ by 2:
$BaCl_2 + 2KOH
ightarrow Ba(OH)_2 + 2KCl$
Check:

• Left: Ba = 1, Cl = 2, K = 2, O = 2, H = 2
• Right: Ba = 1, Cl = 2, K = 2, O = 2, H = 2

Balanced: $\boldsymbol{BaCl_2 + 2KOH
ightarrow Ba(OH)_2 + 2KCl}$

14. $\boldsymbol{H_2O_2

ightarrow H_2O + O_2}$

Step 1: Count atoms

• Left: H = 2, O = 2
• Right: H = 2, O = 1 + 2 = 3

Step 2: Balance O

Multiply $H_2O_2$ by 2, $H_2O$ by 2:
$2H_2O_2
ightarrow 2H_2O + O_2$
Check:

• Left: H = 4, O = 4
• Right: H = 4, O = 2 + 2 = 4

Balanced: $\boldsymbol{2H_2O_2
ightarrow 2H_2O + O_2}$

15. $\boldsymbol{Sb_4O_6 + C

ightarrow Sb + CO}$

Step 1: Count atoms

• Left: Sb = 4, O = 6, C = 1
• Right: Sb = 1, O = 1, C = 1

Step 2: Balance Sb

Multiply $Sb$ by 4: $Sb_4O_6 + C
ightarrow 4Sb + CO$

Step 3: Balance O

Multiply $CO$ by 6: $Sb_4O_6 + C
ightarrow 4Sb + 6CO$

Step 4: Balance C

Multiply $C$ by 6:
$Sb_4O_6 + 6C
ightarrow 4Sb + 6CO$
Check:

• Left: Sb = 4, O = 6, C = 6
• Right: Sb = 4, O = 6, C = 6

Balanced: $\boldsymbol{Sb_4O_6 + 6C
ightarrow 4Sb + 6CO}$

Final Answers (Balanced Equations):

1. $2CaO_2

ightarrow 2CaO + O_2$

1. $H_2 + Cl_2

ightarrow 2HCl$

1. $2Ag_2O

ightarrow 4Ag + O_2$

1. $2Al(OH)_3

ightarrow Al_2O_3 + 3H_2O$

1. $MnO_2 + 4HCl

ightarrow MnCl_2 + 2H_2O + Cl_2$

1. $2MgO_2

ightarrow 2MgO + O_2$

1. $2H_2S + O_2

ightarrow 2S + 2H_2O$

1. $2Al + 3Cl_2

ightarrow 2AlCl_3$

1. $Pb(NO_3)_2 + 2LiCl

ightarrow PbCl_2 + 2LiNO_3$

1. $2Al_2O_3

ightarrow 4Al + 3O_2$

1. $2HgO

ightarrow 2Hg + O_2$

1. $Al(OH)_3 + 3HCl

ightarrow AlCl_3 + 3H_2O$

1. $BaCl_2 + 2KOH

ightarrow Ba(OH)_2 + 2KCl$

1. $2H_2O_2

ightarrow 2H_2O + O_2$

1. $Sb_4O_6 + 6C

ightarrow 4Sb + 6CO$