Question

— c₃h₈ + — o₂ → — co₂ + — h₂o
— na + — cl₂ → — nacl
— h₂o + — co₂ → c₇h₈
— al + — fe₃n₂ → — aln + — fe
— bf₃ + — li₂so₃ → — b₂(so₃)₃ + — lif

Explanation:

First Equation: $\boldsymbol{C_3H_8 + O_2

ightarrow CO_2 + H_2O}$

Step1: Balance Carbon (C)

On the left, $C_3H_8$ has 3 C. So, put 3 in front of $CO_2$:
$C_3H_8 + O_2
ightarrow 3CO_2 + H_2O$

Step2: Balance Hydrogen (H)

$C_3H_8$ has 8 H. So, put 4 in front of $H_2O$ (since $4 \times 2 = 8$):
$C_3H_8 + O_2
ightarrow 3CO_2 + 4H_2O$

Step3: Balance Oxygen (O)

Right side: $3CO_2$ has $3 \times 2 = 6$ O; $4H_2O$ has $4 \times 1 = 4$ O. Total O: $6 + 4 = 10$. So, $O_2$ needs $\frac{10}{2} = 5$:
$C_3H_8 + 5O_2
ightarrow 3CO_2 + 4H_2O$

Second Equation: $\boldsymbol{Na + Cl_2

ightarrow NaCl}$

Step1: Balance Chlorine (Cl)

$Cl_2$ has 2 Cl. So, put 2 in front of $NaCl$:
$Na + Cl_2
ightarrow 2NaCl$

Step2: Balance Sodium (Na)

$2NaCl$ has 2 Na. So, put 2 in front of $Na$:
$2Na + Cl_2
ightarrow 2NaCl$

Third Equation: $\boldsymbol{H_2O + CO_2

ightarrow C_7H_8 + O_2}$ (Assuming Product is $C_7H_8 + O_2$, Likely Combustion)

Step1: Balance Carbon (C)

$C_7H_8$ has 7 C. So, put 7 in front of $CO_2$:
$H_2O + 7CO_2
ightarrow C_7H_8 + O_2$

Step2: Balance Hydrogen (H)

$C_7H_8$ has 8 H. So, put 4 in front of $H_2O$ (since $4 \times 2 = 8$):
$4H_2O + 7CO_2
ightarrow C_7H_8 + O_2$

Step3: Balance Oxygen (O)

Left: $4H_2O$ has $4 \times 1 = 4$ O; $7CO_2$ has $7 \times 2 = 14$ O. Total O: $4 + 14 = 18$.
Right: $O_2$ has $2x$ O; $C_7H_8$ has 0 O. So, $2x = 18 \implies x = 9$:
$4H_2O + 7CO_2
ightarrow C_7H_8 + 9O_2$

Fourth Equation: $\boldsymbol{Al + Fe_3N_2

ightarrow AlN + Fe}$

Answer:

Step1: Balance Boron (B)

$B_2(SO_3)_3$ has 2 B. $BF_3$ has 1 B. So, put 2 in front of $BF_3$:
$2BF_3 + Li_2SO_3
ightarrow B_2(SO_3)_3 + LiF$

Step2: Balance Sulfite ($SO_3$)

$B_2(SO_3)_3$ has 3 $SO_3$. So, put 3 in front of $Li_2SO_3$:
$2BF_3 + 3Li_2SO_3
ightarrow B_2(SO_3)_3 + LiF$

Step3: Balance Lithium (Li)

$3Li_2SO_3$ has 6 Li. $LiF$ has 1 Li. So, put 6 in front of $LiF$:
$2BF_3 + 3Li_2SO_3
ightarrow B_2(SO_3)_3 + 6LiF$

Step4: Balance Fluorine (F)

$2BF_3$ has $2 \times 3 = 6$ F. $6LiF$ has 6 F. Balanced.

Final Balanced Equations:

1. $\boldsymbol{1C_3H_8 + 5O_2

ightarrow 3CO_2 + 4H_2O}$

1. $\boldsymbol{2Na + 1Cl_2

ightarrow 2NaCl}$

1. $\boldsymbol{4H_2O + 7CO_2

ightarrow 1C_7H_8 + 9O_2}$ (Adjusted for Combustion)

1. $\boldsymbol{2Al + 1Fe_3N_2

ightarrow 2AlN + 3Fe}$

1. $\boldsymbol{2BF_3 + 3Li_2SO_3

ightarrow 1B_2(SO_3)_3 + 6LiF}$