balancing equations, type and classification\nname: mod: date\n7. naclo₃ → nacl + o₂\n8. mgcl₂ → mg +…

Explanation:

Step1: Balance $NaClO_3 \to NaCl + O_2$

We need 2 moles of $NaClO_3$ to get 6 oxygen atoms on the left - hand side and 2 moles of $NaCl$ and 3 moles of $O_2$ on the right - hand side. So the balanced equation is $2NaClO_3\rightarrow2NaCl + 3O_2$.

Step2: Balance $MgCl_2\rightarrow Mg+Cl_2$

The equation is already balanced as there is 1 magnesium and 2 chlorine atoms on both sides. So the balanced equation is $MgCl_2\rightarrow Mg + Cl_2$.

Step3: Balance $Fe + S\rightarrow FeS$

The equation is already balanced as there is 1 iron and 1 sulfur atom on both sides. So the balanced equation is $Fe + S\rightarrow FeS$.

Step4: Balance $KClO_3\rightarrow KCl+O_2$

We need 2 moles of $KClO_3$ to get 6 oxygen atoms on the left - hand side and 2 moles of $KCl$ and 3 moles of $O_2$ on the right - hand side. So the balanced equation is $2KClO_3\rightarrow2KCl + 3O_2$.

Step5: Balance $K + H_2O\rightarrow KOH+H_2$

We need 2 moles of $K$, 2 moles of $H_2O$ to get 2 moles of $KOH$ and 1 mole of $H_2$. So the balanced equation is $2K + 2H_2O\rightarrow2KOH + H_2$.

Step6: Balance $Cl_2+LiBr\rightarrow LiCl + Br_2$

We need 2 moles of $LiBr$ and 2 moles of $LiCl$. So the balanced equation is $Cl_2 + 2LiBr\rightarrow2LiCl+Br_2$.

Step7: Balance $Al(NO_3)_3+H_2SO_4\rightarrow Al_2(SO_4)_3 + HNO_3$

We need 2 moles of $Al(NO_3)_3$, 3 moles of $H_2SO_4$ to get 1 mole of $Al_2(SO_4)_3$ and 6 moles of $HNO_3$. So the balanced equation is $2Al(NO_3)_3+3H_2SO_4\rightarrow Al_2(SO_4)_3 + 6HNO_3$.

Answer:

1. $2NaClO_3\rightarrow2NaCl + 3O_2$
2. $MgCl_2\rightarrow Mg + Cl_2$
3. $Fe + S\rightarrow FeS$
4. $2KClO_3\rightarrow2KCl + 3O_2$
5. $2K + 2H_2O\rightarrow2KOH + H_2$
6. $Cl_2 + 2LiBr\rightarrow2LiCl+Br_2$
7. $2Al(NO_3)_3+3H_2SO_4\rightarrow Al_2(SO_4)_3 + 6HNO_3$