Question

balancing chemical equations
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
11. p + o₂ → __ p₂o₅
12. na + h₂o → naoh + h₂
13. ag₂o → ag + __ o₂
14. s₈ + o₂ → __ so₃
15. co₂ + h₂o → c₆h₁₂o₆ + o₂
16. k + mgbr₂ → kbr + mg
17. hcl + caco₃ → cacl₂ + h₂o + __ co₂
18. hno₃ + nahco₃ → nano₃ + h₂o + __ co₂
19. h₂o + o₂ → __ h₂o₂
20. nabr + caf₂ → naf + cabr₂
21. h₂so₄ + nano₂ → hno₂ + na₂so₄

Explanation:

Step1: Balance nitrogen atoms in $N_2 + H_2

ightarrow NH_3$
We have 2 nitrogen atoms on the left - hand side. To balance nitrogen in $NH_3$, we put 2 in front of $NH_3$: $N_2 + H_2
ightarrow 2NH_3$. Then, to balance hydrogen, we put 3 in front of $H_2$. The balanced equation is $N_2+3H_2 = 2NH_3$.

Step2: Balance oxygen atoms in $KClO_3

ightarrow KCl + O_2$
The least - common multiple of 3 (from $KClO_3$) and 2 (from $O_2$) for oxygen atoms is 6. So we put 2 in front of $KClO_3$ and 3 in front of $O_2$: $2KClO_3
ightarrow KCl + 3O_2$. Then we put 2 in front of $KCl$ to balance potassium and chlorine. The balanced equation is $2KClO_3=2KCl + 3O_2$.

Step3: Balance chlorine and fluorine atoms in $NaCl+F_2

ightarrow NaF + Cl_2$
We put 2 in front of $NaCl$ and 2 in front of $NaF$ to balance sodium, chlorine and fluorine. The balanced equation is $2NaCl+F_2 = 2NaF+Cl_2$.

Step4: Balance oxygen atoms in $H_2+O_2

ightarrow H_2O$
We put 2 in front of $H_2O$ to balance oxygen, then 2 in front of $H_2$ to balance hydrogen. The balanced equation is $2H_2+O_2 = 2H_2O$.

Step5: Balance lead and chlorine atoms in $Pb(OH)_2+HCl

ightarrow PbO + PbCl_2+H_2O$
First, rewrite the reaction as two separate reactions for simplicity. But if we consider the overall reaction, we find that this reaction as written may be incorrect or a complex multi - step reaction not in a simple form. Assuming a simple acid - base type reaction, a more likely reaction is $Pb(OH)_2 + 2HCl=PbCl_2+2H_2O$.

Step6: Balance aluminum and sulfate atoms in $AlBr_3+K_2SO_4

ightarrow KBr+Al_2(SO_4)_3$
We put 2 in front of $AlBr_3$ to balance aluminum. Then we put 3 in front of $K_2SO_4$ to balance sulfate and 6 in front of $KBr$ to balance potassium and bromine. The balanced equation is $2AlBr_3 + 3K_2SO_4=6KBr+Al_2(SO_4)_3$.

Step7: Balance carbon and hydrogen atoms in $CH_4+O_2

ightarrow CO_2+H_2O$
We have 1 carbon atom in $CH_4$, so 1 in front of $CO_2$. We have 4 hydrogen atoms in $CH_4$, so 2 in front of $H_2O$. Then we put 2 in front of $O_2$ to balance oxygen. The balanced equation is $CH_4+2O_2 = CO_2+2H_2O$.

Step8: Balance carbon and hydrogen atoms in $C_3H_8+O_2

ightarrow CO_2+H_2O$
We put 3 in front of $CO_2$ to balance carbon and 4 in front of $H_2O$ to balance hydrogen. Then we put 5 in front of $O_2$ to balance oxygen. The balanced equation is $C_3H_8+5O_2 = 3CO_2+4H_2O$.

Step9: Balance carbon and hydrogen atoms in $C_8H_{18}+O_2

ightarrow CO_2+H_2O$
We put 8 in front of $CO_2$ to balance carbon and 9 in front of $H_2O$ to balance hydrogen. Then we put $\frac{25}{2}$ in front of $O_2$ to balance oxygen. To get whole - number coefficients, we multiply all coefficients by 2. The balanced equation is $2C_8H_{18}+25O_2 = 16CO_2+18H_2O$.

Step10: Balance iron and chlorine atoms in $FeCl_3+NaOH

ightarrow Fe(OH)_3+NaCl$
We put 3 in front of $NaOH$ and 3 in front of $NaCl$ to balance sodium, chlorine and hydroxide. The balanced equation is $FeCl_3+3NaOH = Fe(OH)_3+3NaCl$.

Step11: Balance phosphorus and oxygen atoms in $P+O_2

ightarrow P_2O_5$
We put 2 in front of $P$ to balance phosphorus and $\frac{5}{2}$ in front of $O_2$ to balance oxygen. To get whole - number coefficients, we multiply all coefficients by 2. The balanced equation is $4P + 5O_2=2P_2O_5$.

Step12: Balance sodium and hydrogen atoms in $Na+H_2O

ightarrow NaOH+H_2$
We put 2 in front of $Na$ and 2 in front of $H_2O$ to balance sodium, hydrogen and oxygen. The balanced equation is $2Na+2H_2O = 2NaOH+H_2$.

Step13: Balance silver and oxygen atoms in $Ag_2O

ightarrow Ag+O_2$
We put 2 in front of $Ag_2O$ to balance oxygen, then 4 in front of $Ag$ to balance silver. The balanced equation is $2Ag_2O=4Ag + O_2$.

Step14: Balance sulfur and oxygen atoms in $S_8+O_2

ightarrow SO_3$
We put 8 in front of $SO_3$ to balance sulfur, then 12 in front of $O_2$ to balance oxygen. The balanced equation is $S_8+12O_2 = 8SO_3$.

Step15: Balance carbon, hydrogen and oxygen atoms in $CO_2+H_2O

ightarrow C_6H_{12}O_6+O_2$
We put 6 in front of $CO_2$ and 6 in front of $H_2O$ to balance carbon and hydrogen. Then we put 6 in front of $O_2$ to balance oxygen. The balanced equation is $6CO_2+6H_2O = C_6H_{12}O_6+6O_2$.

Step16: Balance potassium, magnesium and bromine atoms in $K+MgBr

ightarrow KBr+Mg$
The correct formula for magnesium bromide is $MgBr_2$. The balanced equation for $K + MgBr_2
ightarrow KBr+Mg$ is $2K+MgBr_2 = 2KBr+Mg$.

Step17: Balance hydrogen, chlorine, calcium, carbon and oxygen atoms in $HCl+CaCO_3

ightarrow CaCl_2+H_2O+CO_2$
We put 2 in front of $HCl$ to balance chlorine and hydrogen. The balanced equation is $2HCl+CaCO_3 = CaCl_2+H_2O+CO_2$.

Step18: Balance hydrogen, nitrogen, sodium, carbon and oxygen atoms in $HNO_3+NaHCO_3

ightarrow NaNO_3+H_2O+CO_2$
The equation is already balanced as written: $HNO_3+NaHCO_3 = NaNO_3+H_2O+CO_2$.

Step19: Balance hydrogen and oxygen atoms in $H_2O+O_2

ightarrow H_2O_2$
We put 2 in front of $H_2O$ and 2 in front of $H_2O_2$ to balance hydrogen and oxygen. The balanced equation is $2H_2O+O_2 = 2H_2O_2$.

Step20: Balance sodium, bromine, calcium and fluorine atoms in $NaBr+CaF_2

ightarrow NaF+CaBr_2$
We put 2 in front of $NaBr$ and 2 in front of $NaF$ to balance sodium, bromine and fluorine. The balanced equation is $2NaBr+CaF_2 = 2NaF+CaBr_2$.

Step21: Balance hydrogen, sulfur, sodium, nitrogen and oxygen atoms in $H_2SO_4+NaNO_2

ightarrow HNO_2+Na_2SO_4$
We put 2 in front of $NaNO_2$ and 2 in front of $HNO_2$ to balance sodium, nitrogen, hydrogen and oxygen. The balanced equation is $H_2SO_4+2NaNO_2 = 2HNO_2+Na_2SO_4$.

Answer:

1. $N_2+3H_2 = 2NH_3$
2. $2KClO_3=2KCl + 3O_2$
3. $2NaCl+F_2 = 2NaF+Cl_2$
4. $2H_2+O_2 = 2H_2O$
5. $Pb(OH)_2 + 2HCl=PbCl_2+2H_2O$
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. $FeCl_3+3NaOH = Fe(OH)_3+3NaCl$
11. $4P + 5O_2=2P_2O_5$
12. $2Na+2H_2O = 2NaOH+H_2$
13. $2Ag_2O=4Ag + O_2$
14. $S_8+12O_2 = 8SO_3$
15. $6CO_2+6H_2O = C_6H_{12}O_6+6O_2$
16. $2K+MgBr_2 = 2KBr+Mg$
17. $2HCl+CaCO_3 = CaCl_2+H_2O+CO_2$
18. $HNO_3+NaHCO_3 = NaNO_3+H_2O+CO_2$
19. $2H_2O+O_2 = 2H_2O_2$
20. $2NaBr+CaF_2 = 2NaF+CaBr_2$
21. $H_2SO_4+2NaNO_2 = 2HNO_2+Na_2SO_4$