balance the equations below:
1) ____ n₂ + ____ h₂ → ____ nh₃
2) ____ kclo₃ → ____ kcl + ____ o₂
3) ____ nacl + ____ f₂ → ____ naf + ____ cl₂
4) ____ h₂ + ____ o₂ → ____ h₂o
5) ____ pb(oh)₂ + ____ hcl → ____ h₂o + ____ pbcl₂
6) ____ albr₃ + ____ k₂so₄ → ____ kbr + ____ al₂(so₄)₃
7) ____ ch₄ + ____ o₂ → ____ co₂ + ____ h₂o
8) ____ c₃h₈ + ____ o₂ → ____ co₂ + ____ h₂o
9) ____ c₈h₁₈ + ____ o₂ → ____ co₂ + ____ h₂o
×10) ____ fecl₃ + ____ naoh → ____ fe(oh)₃ + ____ nacl

Question

Accepted Answer

# Explanation:
## Step1: Balance the first equation
For $N_2 + H_2
ightarrow NH_3$, there are 2 $N$ atoms on the left. So put 2 in front of $NH_3$. Then there are 6 $H$ atoms on the right, so put 3 in front of $H_2$.
$N_2+3H_2 = 2NH_3$
## Step2: Balance the second equation
For $KClO_3
ightarrow KCl + O_2$, there are 3 $O$ atoms in $KClO_3$ and 2 in $O_2$. The least - common multiple of 3 and 2 is 6. Put 2 in front of $KClO_3$ and 3 in front of $O_2$. Then there are 2 $K$ and 2 $Cl$ atoms, so put 2 in front of $KCl$.
$2KClO_3=2KCl + 3O_2$
## Step3: Balance the third equation
For $NaCl+F_2
ightarrow NaF + Cl_2$, there are 2 $Cl$ atoms in $Cl_2$ and 2 $F$ atoms in $F_2$. Put 2 in front of $NaCl$ and 2 in front of $NaF$.
$2NaCl+F_2 = 2NaF+Cl_2$
## Step4: Balance the fourth equation
For $H_2+O_2
ightarrow H_2O$, there are 2 $O$ atoms on the left and 1 on the right. Put 2 in front of $H_2O$, then there are 4 $H$ atoms, so put 2 in front of $H_2$.
$2H_2+O_2 = 2H_2O$
## Step5: Balance the fifth equation
For $Pb(OH)_2+HCl
ightarrow H_2O+PbCl_2$, there are 2 $Cl$ atoms in $PbCl_2$, so put 2 in front of $HCl$. Then there are 4 $H$ atoms on the left ($2$ from $Pb(OH)_2$ and $2$ from $HCl$) and 2 on the right in $H_2O$, so put 2 in front of $H_2O$.
$Pb(OH)_2 + 2HCl=2H_2O+PbCl_2$
## Step6: Balance the sixth equation
For $AlBr_3+K_2SO_4
ightarrow KBr+Al_2(SO_4)_3$, there are 2 $Al$ atoms in $Al_2(SO_4)_3$, so put 2 in front of $AlBr_3$. There are 3 $SO_4^{2 -}$ groups in $Al_2(SO_4)_3$, so put 3 in front of $K_2SO_4$. Then there are 6 $K$ atoms, so put 6 in front of $KBr$.
$2AlBr_3+3K_2SO_4 = 6KBr+Al_2(SO_4)_3$
## Step7: Balance the seventh equation
For $CH_4+O_2
ightarrow CO_2+H_2O$, there is 1 $C$ atom. For $H$, there are 4 in $CH_4$, so put 2 in front of $H_2O$. Then there are 4 $O$ atoms in $CO_2$ and $2H_2O$ (total 4 $O$ atoms from $CO_2$ and $2H_2O$), so put 2 in front of $O_2$.
$CH_4 + 2O_2=CO_2+2H_2O$
## Step8: Balance the eighth equation
For $C_3H_8+O_2
ightarrow CO_2+H_2O$, there are 3 $C$ atoms. So put 3 in front of $CO_2$. There are 8 $H$ atoms, so put 4 in front of $H_2O$. Then the total number of $O$ atoms is $3\times2 + 4\times1=10$, so put 5 in front of $O_2$.
$C_3H_8+5O_2 = 3CO_2+4H_2O$
## Step9: Balance the ninth equation
For $C_8H_{18}+O_2
ightarrow CO_2+H_2O$, there are 8 $C$ atoms. So put 8 in front of $CO_2$. There are 18 $H$ atoms, so put 9 in front of $H_2O$. The total number of $O$ atoms is $8\times2+9\times1 = 25$, so put $\frac{25}{2}$ in front of $O_2$. Multiply through by 2 to get rid of the fraction: $2C_8H_{18}+25O_2 = 16CO_2+18H_2O$
## Step10: Balance the tenth equation
For $FeCl_3+NaOH
ightarrow Fe(OH)_3+NaCl$, there are 3 $Cl$ atoms in $FeCl_3$, so put 3 in front of $NaCl$. There are 3 $OH^-$ groups in $Fe(OH)_3$, so put 3 in front of $NaOH$.
$FeCl_3+3NaOH = Fe(OH)_3+3NaCl$

# Answer:
1. $1N_2 + 3H_2=2NH_3$
2. $2KClO_3=2KCl + 3O_2$
3. $2NaCl+F_2 = 2NaF+Cl_2$
4. $2H_2+O_2 = 2H_2O$
5. $Pb(OH)_2 + 2HCl=2H_2O+PbCl_2$
6. $2AlBr_3+3K_2SO_4 = 6KBr+Al_2(SO_4)_3$
7. $CH_4 + 2O_2=CO_2+2H_2O$
8. $C_3H_8+5O_2 = 3CO_2+4H_2O$
9. $2C_8H_{18}+25O_2 = 16CO_2+18H_2O$
10. $FeCl_3+3NaOH = Fe(OH)_3+3NaCl$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Balance the first equation

Step2: Balance the second equation

Step3: Balance the third equation

Step4: Balance the fourth equation

Step5: Balance the fifth equation

Step6: Balance the sixth equation

Step7: Balance the seventh equation

Step8: Balance the eighth equation

Step9: Balance the ninth equation

Step10: Balance the tenth equation

Answer: