1. balance each equation for a synthesis reaction.\n(a) k + o₂ → k₂o\n(b) p₄ + cl₂ → pcl₅\n(c) cu + s₈ →…

Explanation:

Step1: Balance equation (a)

Count atoms on both sides. For $K$ and $O$, we need 4 potassium atoms on the left - hand side to balance the oxygen in $K_2O$. So the balanced equation is $4K+O_2\rightarrow 2K_2O$.

Step2: Balance equation (b)

On the left - hand side, we have 4 phosphorus atoms in $P_4$. To balance the chlorine, we need 10 moles of $Cl_2$ and 4 moles of $PCl_5$. The balanced equation is $P_4 + 10Cl_2\rightarrow 4PCl_5$.

Step3: Balance equation (c)

There are 8 sulfur atoms in $S_8$. We need 8 moles of $Cu$ and 8 moles of $CuS$. The balanced equation is $8Cu+S_8\rightarrow 8CuS$.

Step4: Balance equation (d)

To balance the oxygen in $MgO$ and magnesium, we need 2 moles of $Mg$ and 2 moles of $MgO$. The balanced equation is $2Mg + O_2\rightarrow 2MgO$.

Step5: Balance equation (e)

For iron and oxygen in $Fe_2O_3$, we need 4 moles of $Fe$ and 3 moles of $O_2$. The balanced equation is $4Fe+3O_2\rightarrow 2Fe_2O_3$.

Step6: Balance equation (f)

There are 4 phosphorus atoms in $P_4$ and 8 sulfur atoms in $S_8$. The balanced equation is $2P_4+5S_8\rightarrow 8P_2S_5$.

Step7: Balance equation (g)

To balance carbon and oxygen in $CO$, we need 2 moles of $C$ and 2 moles of $CO$. The balanced equation is $2C+O_2\rightarrow 2CO$.

Step8: Balance equation (h)

For nitrogen and oxygen in $NO_2$, we need 1 mole of $N_2$ and 2 moles of $O_2$. The balanced equation is $N_2 + 2O_2\rightarrow 2NO_2$.

Step9: Balance equation (i)

To balance lithium and nitrogen in $LiN$, we need 6 moles of $Li$ and 2 moles of $LiN$. The balanced equation is $6Li+N_2\rightarrow 2Li_3N$.

Answer:

(a) $4K+O_2\rightarrow 2K_2O$ (b) $P_4 + 10Cl_2\rightarrow 4PCl_5$ (c) $8Cu+S_8\rightarrow 8CuS$ (d) $2Mg + O_2\rightarrow 2MgO$ (e) $4Fe+3O_2\rightarrow 2Fe_2O_3$ (f) $2P_4+5S_8\rightarrow 8P_2S_5$ (g) $2C+O_2\rightarrow 2CO$ (h) $N_2 + 2O_2\rightarrow 2NO_2$ (i) $6Li+N_2\rightarrow 2Li_3N$