Question

final final equation
h-1 h-1 2 h₂ + o₂ → 2 h₂o
o-2 o-2
h-1 h-1 2 h₂o₂ → 2 h₂o + o₂
o-4 o-4
(some unclear or partial entries)
4 na + o₂ = 2 na₂o
n₂ + 3 h₂ = 2 nh₃
p₄ + 5 o₂ = p₄o₁₀
3 fe + 4 h₂o = (partial, maybe fe₃o₄?)
c + 2 h₂ = ch₄
na₂so₄ + cacl₂ = caso₄ + 2 nacl
2 c₂h₆ + 7 o₂ = 4 (partial, maybe co₂) + 6 h₂o
2 al₂o₃ = 4 al + (partial, maybe 3 o₂)

Explanation:

To balance the chemical equations, we'll follow the steps of counting atoms on each side and adjusting coefficients. Let's take one equation at a time (e.g., $\boldsymbol{2H_2 + O_2
ightarrow 2H_2O}$ is already balanced, let's do $\boldsymbol{N_2 + 3H_2
ightarrow 2NH_3}$):

Step 1: Count Atoms (Unbalanced)

• Left (Reactants): $N$: 2, $H$: $3\times2 = 6$
• Right (Products): $N$: $2\times1 = 2$, $H$: $2\times3 = 6$

Wait, this one is balanced! Let's try $\boldsymbol{P_4 + 5O_2
ightarrow P_4O_{10}}$:

Step 1: Count Atoms (Unbalanced)

• Left: $P$: 4, $O$: $5\times2 = 10$
• Right: $P$: 4, $O$: 10

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

Step 1: Count Atoms (Unbalanced)

• Left: $H$: $2\times2 = 4$, $O$: $2\times2 = 4$
• Right: $H$: $2\times2 = 4$, $O$: $(2\times1) + 2 = 4$

Balanced! For $\boldsymbol{4Na + O_2
ightarrow 2Na_2O}$:

Step 1: Count Atoms (Unbalanced)

• Left: $Na$: 4, $O$: 2
• Right: $Na$: $2\times2 = 4$, $O$: $2\times1 = 2$

Balanced! For $\boldsymbol{3Fe + 4H_2O
ightarrow Fe_3O_4 + 4H_2}$ (assuming the original was incomplete, let's correct):

Step 1: Count Atoms (Unbalanced)

• Left: $Fe$: 3, $H$: $4\times2 = 8$, $O$: 4
• Right: $Fe$: 3, $H$: $4\times2 = 8$, $O$: 4

Balanced! For $\boldsymbol{C + 2H_2
ightarrow CH_4}$:

Step 1: Count Atoms (Unbalanced)

• Left: $C$: 1, $H$: $2\times2 = 4$
• Right: $C$: 1, $H$: 4

Balanced! For $\boldsymbol{Na_2SO_4 + CaCl_2
ightarrow CaSO_4 + 2NaCl}$:

Step 1: Count Atoms (Unbalanced)

• Left: $Na$: 2, $S$: 1, $O$: 4, $Ca$: 1, $Cl$: 2
• Right: $Na$: $2\times1 = 2$, $S$: 1, $O$: 4, $Ca$: 1, $Cl$: $2\times1 = 2$

Balanced! For $\boldsymbol{2C_2H_6 + 7O_2
ightarrow 4CO_2 + 6H_2O}$:

Step 1: Count Atoms (Unbalanced)

• Left: $C$: $2\times2 = 4$, $H$: $2\times6 = 12$, $O$: $7\times2 = 14$
• Right: $C$: $4\times1 = 4$, $H$: $6\times2 = 12$, $O$: $(4\times2) + 6 = 14$

Balanced! For $\boldsymbol{2Al_2O_3
ightarrow 4Al + 3O_2}$ (correcting the original):

Step 1: Count Atoms (Unbalanced)

• Left: $Al$: $2\times2 = 4$, $O$: $2\times3 = 6$
• Right: $Al$: 4, $O$: $3\times2 = 6$

Balanced!

Final Balanced Equations (Selected Examples):

1. $2H_2 + O_2

ightarrow 2H_2O$ (balanced)

1. $2H_2O_2

ightarrow 2H_2O + O_2$ (balanced)

1. $4Na + O_2

ightarrow 2Na_2O$ (balanced)

1. $N_2 + 3H_2

ightarrow 2NH_3$ (balanced)

1. $P_4 + 5O_2

ightarrow P_4O_{10}$ (balanced)

1. $3Fe + 4H_2O

ightarrow Fe_3O_4 + 4H_2$ (balanced)

1. $C + 2H_2

ightarrow CH_4$ (balanced)

1. $Na_2SO_4 + CaCl_2

ightarrow CaSO_4 + 2NaCl$ (balanced)

1. $2C_2H_6 + 7O_2

ightarrow 4CO_2 + 6H_2O$ (balanced)

1. $2Al_2O_3

ightarrow 4Al + 3O_2$ (balanced)

(Note: Some original equations had typos; corrections were made for clarity and balance.)

Answer:

To balance the chemical equations, we'll follow the steps of counting atoms on each side and adjusting coefficients. Let's take one equation at a time (e.g., $\boldsymbol{2H_2 + O_2
ightarrow 2H_2O}$ is already balanced, let's do $\boldsymbol{N_2 + 3H_2
ightarrow 2NH_3}$):

Step 1: Count Atoms (Unbalanced)

• Left (Reactants): $N$: 2, $H$: $3\times2 = 6$
• Right (Products): $N$: $2\times1 = 2$, $H$: $2\times3 = 6$

Wait, this one is balanced! Let's try $\boldsymbol{P_4 + 5O_2
ightarrow P_4O_{10}}$:

Step 1: Count Atoms (Unbalanced)

• Left: $P$: 4, $O$: $5\times2 = 10$
• Right: $P$: 4, $O$: 10

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

Step 1: Count Atoms (Unbalanced)

• Left: $H$: $2\times2 = 4$, $O$: $2\times2 = 4$
• Right: $H$: $2\times2 = 4$, $O$: $(2\times1) + 2 = 4$

Balanced! For $\boldsymbol{4Na + O_2
ightarrow 2Na_2O}$:

Step 1: Count Atoms (Unbalanced)

• Left: $Na$: 4, $O$: 2
• Right: $Na$: $2\times2 = 4$, $O$: $2\times1 = 2$

Balanced! For $\boldsymbol{3Fe + 4H_2O
ightarrow Fe_3O_4 + 4H_2}$ (assuming the original was incomplete, let's correct):

Step 1: Count Atoms (Unbalanced)

• Left: $Fe$: 3, $H$: $4\times2 = 8$, $O$: 4
• Right: $Fe$: 3, $H$: $4\times2 = 8$, $O$: 4

Balanced! For $\boldsymbol{C + 2H_2
ightarrow CH_4}$:

Step 1: Count Atoms (Unbalanced)

• Left: $C$: 1, $H$: $2\times2 = 4$
• Right: $C$: 1, $H$: 4

Balanced! For $\boldsymbol{Na_2SO_4 + CaCl_2
ightarrow CaSO_4 + 2NaCl}$:

Step 1: Count Atoms (Unbalanced)

• Left: $Na$: 2, $S$: 1, $O$: 4, $Ca$: 1, $Cl$: 2
• Right: $Na$: $2\times1 = 2$, $S$: 1, $O$: 4, $Ca$: 1, $Cl$: $2\times1 = 2$

Balanced! For $\boldsymbol{2C_2H_6 + 7O_2
ightarrow 4CO_2 + 6H_2O}$:

Step 1: Count Atoms (Unbalanced)

• Left: $C$: $2\times2 = 4$, $H$: $2\times6 = 12$, $O$: $7\times2 = 14$
• Right: $C$: $4\times1 = 4$, $H$: $6\times2 = 12$, $O$: $(4\times2) + 6 = 14$

Balanced! For $\boldsymbol{2Al_2O_3
ightarrow 4Al + 3O_2}$ (correcting the original):

Step 1: Count Atoms (Unbalanced)

• Left: $Al$: $2\times2 = 4$, $O$: $2\times3 = 6$
• Right: $Al$: 4, $O$: $3\times2 = 6$

Balanced!

Final Balanced Equations (Selected Examples):

1. $2H_2 + O_2

ightarrow 2H_2O$ (balanced)

1. $2H_2O_2

ightarrow 2H_2O + O_2$ (balanced)

1. $4Na + O_2

ightarrow 2Na_2O$ (balanced)

1. $N_2 + 3H_2

ightarrow 2NH_3$ (balanced)

1. $P_4 + 5O_2

ightarrow P_4O_{10}$ (balanced)

1. $3Fe + 4H_2O

ightarrow Fe_3O_4 + 4H_2$ (balanced)

1. $C + 2H_2

ightarrow CH_4$ (balanced)

1. $Na_2SO_4 + CaCl_2

ightarrow CaSO_4 + 2NaCl$ (balanced)

1. $2C_2H_6 + 7O_2

ightarrow 4CO_2 + 6H_2O$ (balanced)

1. $2Al_2O_3

ightarrow 4Al + 3O_2$ (balanced)

(Note: Some original equations had typos; corrections were made for clarity and balance.)