1. 2k + cl₂ → 2kcl answer: 2,1,2\n2. n₂ + h₂ → nh₃\n3. naf + br₂ → nabr + f₂\n4. so₂ + li₂se → sse₂ +…

Explanation:

Step1: Balance nitrogen in $N_2 + H_2\rightarrow NH_3$

There are 2 nitrogen atoms on the left - hand side. To balance nitrogen, put 2 in front of $NH_3$. So the equation becomes $N_2 + H_2\rightarrow2NH_3$.

Step2: Balance hydrogen in $N_2 + H_2\rightarrow2NH_3$

Now, there are 6 hydrogen atoms on the right - hand side. So put 3 in front of $H_2$. The balanced equation is $N_2+3H_2\rightarrow2NH_3$. Answer: 1, 3, 2

Step3: Balance $NaF + Br_2\rightarrow NaBr+F_2$

For sodium and bromine, put 2 in front of $NaF$ and $NaBr$. The balanced equation is $2NaF + Br_2\rightarrow2NaBr+F_2$. Answer: 2, 1, 2, 1

Step4: Balance $SO_2 + Li_2Se\rightarrow SSe_2+Li_2O$

For oxygen, put 2 in front of $Li_2O$. Then for lithium, put 4 in front of $Li_2Se$. For selenium, put 2 in front of $SSe_2$. The balanced equation is $SO_2 + 4Li_2Se\rightarrow2SSe_2+4Li_2O$. Answer: 1, 4, 2, 4

Step5: Balance $PbSO_4\rightarrow PbSO_3+O_2$

Put 2 in front of $PbSO_4$ and $PbSO_3$ to balance lead, sulfur and oxygen. The balanced equation is $2PbSO_4\rightarrow2PbSO_3 + O_2$. Answer: 2, 2, 1

Step6: Balance $Al+Fe_2O_3\rightarrow Fe + Al_2O_3$

For aluminum, put 2 in front of $Al$. For iron, put 2 in front of $Fe$. The balanced equation is $2Al+Fe_2O_3\rightarrow2Fe + Al_2O_3$. Answer: 2, 1, 2, 1

Step7: Balance $SF_6\rightarrow S+F_2$

There are 6 fluorine atoms on the left - hand side. So put 3 in front of $F_2$. The balanced equation is $SF_6\rightarrow S + 3F_2$. Answer: 1, 1, 3

Step8: Balance $C_2H_4+O_2\rightarrow CO_2+H_2O$

For carbon, put 2 in front of $CO_2$. For hydrogen, there are 4 hydrogen atoms in $C_2H_4$, so put 2 in front of $H_2O$. Then for oxygen, there are 6 oxygen atoms on the right - hand side, so put 3 in front of $O_2$. The balanced equation is $C_2H_4+3O_2\rightarrow2CO_2 + 2H_2O$. Answer: 1, 3, 2, 2

Step9: Balance $KClO_3\rightarrow KCl+O_2$

For oxygen, 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 put 2 in front of $KCl$. The balanced equation is $2KClO_3\rightarrow2KCl+3O_2$. Answer: 2, 2, 3

Step10: Balance $Pb(NO_3)_2+NaCl\rightarrow NaNO_3+PbCl_2$

For chlorine, put 2 in front of $NaCl$. For sodium and nitrate, put 2 in front of $NaNO_3$. The balanced equation is $Pb(NO_3)_2 + 2NaCl\rightarrow2NaNO_3+PbCl_2$. Answer: 1, 2, 2, 1

Answer:

2. 1, 3, 2
3. 2, 1, 2, 1
4. 1, 4, 2, 4
5. 2, 2, 1
6. 2, 1, 2, 1
7. 1, 1, 3
8. 1, 3, 2, 2
9. 2, 2, 3
10. 1, 2, 2, 1