balance the equations below. show your work.
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) ___ hcl + ___ caco₃ → ___ cacl₂ + ___ h₂o + ___ co₂
8) ___ ch₄ + ___ o₂ → ___ co₂ + ___ h₂o
9) ___ k + ___ mgbr₂ → ___ kbr + ___ mg
10) ___ fecl₃ + ___ naoh → ___ fe(oh)₃ + ___ nacl
11) ___ co₂ + ___ h₂o → ___ c₆h₁₂o₆ + ___ o₂
12) ___ na + ___ h₂o → ___ naoh + ___ h₂
13) ___ ag₂o → ___ ag + ___ o₂
14) ___ s₈ + ___ o₂ → ___ so₃

Question

Accepted Answer

# Explanation:
## Step1: Balance the first equation
For $N_2 + H_2
ightarrow NH_3$, balance $N$ atoms first. There are $2$ $N$ atoms on the left. So, put $2$ in front of $NH_3$. Now, for $H$ atoms, there are $6$ $H$ atoms on the right. So, put $3$ in front of $H_2$. The balanced equation is $N_2+3H_2 = 2NH_3$.
## Step2: Balance the second equation
For $KClO_3
ightarrow KCl + O_2$, balance $O$ atoms. The least - common multiple of $3$ and $2$ (the number of $O$ atoms in $KClO_3$ and $O_2$ respectively) is $6$. Put $2$ in front of $KClO_3$ and $3$ in front of $O_2$. Then, since there are $2$ $K$ and $2$ $Cl$ atoms from $2KClO_3$, put $2$ in front of $KCl$. The balanced equation is $2KClO_3=2KCl + 3O_2$.
## Step3: Balance the third equation
For $NaCl+F_2
ightarrow NaF + Cl_2$, balance $Cl$ and $F$ atoms. 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 $NaF$. The balanced equation is $2NaCl+F_2 = 2NaF+Cl_2$.
## Step4: Balance the fourth equation
For $H_2+O_2
ightarrow H_2O$, balance $O$ atoms. Put $2$ in front of $H_2O$ to get $2$ $O$ atoms. Then, since there are $4$ $H$ atoms from $2H_2O$, put $2$ in front of $H_2$. The balanced equation is $2H_2+O_2 = 2H_2O$.
## Step5: Balance the fifth equation
For $Pb(OH)_2+HCl
ightarrow H_2O+PbCl_2$, balance $Cl$ atoms. Put $2$ in front of $HCl$. Then, for $H$ and $O$ atoms, since $Pb(OH)_2$ provides $2$ $O$ and $2$ $H$ atoms and $2HCl$ provides $2$ $H$ atoms, the $H_2O$ coefficient is $2$. The balanced equation is $Pb(OH)_2 + 2HCl=2H_2O+PbCl_2$.
## Step6: Balance the sixth equation
For $AlBr_3+K_2SO_4
ightarrow KBr+Al_2(SO_4)_3$, balance $Al$ atoms first. Put $2$ in front of $AlBr_3$. Then, balance $SO_4$ groups. Put $3$ in front of $K_2SO_4$. Since there are $6$ $K$ atoms from $3K_2SO_4$, put $6$ in front of $KBr$. The balanced equation is $2AlBr_3+3K_2SO_4 = 6KBr+Al_2(SO_4)_3$.
## Step7: Balance the seventh equation
For $HCl+CaCO_3
ightarrow CaCl_2+H_2O+CO_2$, balance $Cl$ atoms. Put $2$ in front of $HCl$. Then, check other atoms. The balanced equation is $2HCl+CaCO_3=CaCl_2+H_2O+CO_2$.
## Step8: Balance the eighth equation
For $CH_4+O_2
ightarrow CO_2+H_2O$, balance $C$ atoms (already balanced). Balance $H$ atoms: put $2$ in front of $H_2O$. Then, for $O$ atoms, since there are $4$ $O$ atoms from $2H_2O$ and $2$ $O$ atoms from $CO_2$ (total $4$ $O$ atoms), put $2$ in front of $O_2$. The balanced equation is $CH_4 + 2O_2=CO_2+2H_2O$.
## Step9: Balance the ninth equation
For $K+MgBr_2
ightarrow KBr+Mg$, balance $Br$ atoms. Put $2$ in front of $KBr$. Then, since there are $2$ $K$ atoms from $2KBr$, put $2$ in front of $K$. The balanced equation is $2K+MgBr_2 = 2KBr+Mg$.
## Step10: Balance the tenth equation
For $FeCl_3+NaOH
ightarrow Fe(OH)_3+NaCl$, balance $Cl$ atoms. Put $3$ in front of $NaCl$. Then, for $Na$ atoms, since there are $3$ $Na$ atoms from $3NaCl$, put $3$ in front of $NaOH$. The balanced equation is $FeCl_3+3NaOH = Fe(OH)_3+3NaCl$.
## Step11: Balance the eleventh equation
For $CO_2+H_2O
ightarrow C_6H_{12}O_6+O_2$, balance $C$ atoms. Put $6$ in front of $CO_2$. Then, for $H$ atoms, since there are $12$ $H$ atoms from $C_6H_{12}O_6$, put $6$ in front of $H_2O$. For $O$ atoms, on the left - hand side, there are $6\times2 + 6\times1=18$ $O$ atoms. On the right - hand side, $C_6H_{12}O_6$ has $6$ $O$ atoms. So, $O_2$ should have $\frac{18 - 6}{2}=6$ molecules. The balanced equation is $6CO_2+6H_2O = C_6H_{12}O_6+6O_2$.
## Step12: Balance the twelfth equation
For $Na+H_2O
ightarrow NaOH+H_2$, balance $H$ atoms. Put $2$ in front of $H_2O$ and $2$ in front of $NaOH$. Then, since there are $2$ $Na$ atoms from $2NaOH$, put $2$ in front of $Na$. The balanced equation is $2Na+2H_2O = 2NaOH+H_2$.
## Step13: Balance the thirteenth equation
For $Ag_2O
ightarrow Ag+O_2$, balance $O$ atoms. Put $2$ in front of $Ag_2O$ and $1$ in front of $O_2$. Then, since there are $4$ $Ag$ atoms from $2Ag_2O$, put $4$ in front of $Ag$. The balanced equation is $2Ag_2O=4Ag + O_2$.
## Step14: Balance the fourteenth equation
For $S_8+O_2
ightarrow SO_3$, balance $S$ atoms. Put $8$ in front of $SO_3$. Then, for $O$ atoms, since there are $24$ $O$ atoms from $8SO_3$, put $12$ in front of $O_2$. The balanced equation is $S_8+12O_2 = 8SO_3$.

# 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. $2HCl+CaCO_3=CaCl_2+H_2O+CO_2$
8. $CH_4 + 2O_2=CO_2+2H_2O$
9. $2K+MgBr_2 = 2KBr+Mg$
10. $FeCl_3+3NaOH = Fe(OH)_3+3NaCl$
11. $6CO_2+6H_2O = C_6H_{12}O_6+6O_2$
12. $2Na+2H_2O = 2NaOH+H_2$
13. $2Ag_2O=4Ag + O_2$
14. $S_8+12O_2 = 8SO_3$

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

Step11: Balance the eleventh equation

Answer: