## Chemistry 10th Edition

(a) $4Na + O_2 -- \gt 2Na_2O$ (b) $Mg_3N_2 + 6H_2O -- \gt 2NH_3 + 3Mg(OH)_2$ (c) $2LiCl + Pb(NO_3)_2 -- \gt PbCl_2 + 2LiNO_3$ (d) $2H_2O + 4KO_2 -- \gt 4KOH + 3O_2$ (e) $H_2SO_4 + 2NH_3 -- \gt (NH_4)_2SO_4$
(a) 1. Start by balancing the kind of atoms that appear only one time in each side: $Na$ and $O$ - Balance $O$ The subscript for oxygen on the reactants side is "2", from $O_2$, so we are going to put a "2" as the coefficient of the product with oxygen: $Na + O_2 -- \gt 2Na_2O$ - Balance $Na$: Now, the products side has 2 molecules of $Na_2O$, and each one has 2 sodium atoms, so we have a total of 4 $Na$. To balance the equation, we should put a "$4$" as the coefficient of $Na$ from the reactants. $4Na + O_2 -- \gt 2Na_2O$ - The equation is balanced (b) 1. Start by balancing the kind of atoms that appear only one time in each side: $N$, $Mg$ and $O$. - Balance $Mg$: The subscript for magnesium on the reactants side is "3", so we should put this number as the coefficient of "$Mg(OH)_2$": $Mg_3N_2 + H_2O -- \gt NH_3 + 3Mg(OH)_2$ - Balance $N$: The subscript for nitrogen on the reactants side is "2", so we should put this number as the coefficient of "$NH_3$": $Mg_3N_2 + H_2O -- \gt 2NH_3 + 3Mg(OH)_2$ - Balance $O$: - There is a total of 6 oxygen atoms in $3Mg(OH)_2$, so we should put a 6 as the coefficient of $H_2O$: $Mg_3N_2 + 6H_2O -- \gt 2NH_3 + 3Mg(OH)_2$ - The equation is balanced (c)1. Start by balancing the kind of atoms that appear only one time in each side: $Li$, $Pb$, $N$, $O$ and $Cl$. - Balance $Cl$: The subscript for chlorine on the products side is "2", so we should put this number as the coefficient of "$LiCl$": $2LiCl + Pb(NO_3)_2 -- \gt PbCl_2 + LiNO_3$ - Balance $N$: The reactants side has 2 nitrogens in $Pb(NO_3)_2$, so, to balance that, we should put a "2" as the coefficient of $LiNO_3$. $2LiCl + Pb(NO_3)_2 -- \gt PbCl_2 + 2LiNO_3$ - The equation is balanced (d) 1. Start by balancing the kind of atoms that appear only one time in each side: $K$ and $H$: -Balance $H$: The subscript for hydrogen on the reactants side is "2", so we should put this number as the coefficient of "$KOH$": $H_2O + KO_2 -- \gt 2KOH + O_2$ - Balance $K$: We have 2 potassium on the products side, therefore, to balance the equation, we should put a "2" on the $KO_2$: $H_2O + 2KO_2 -- \gt 2KOH + O_2$ - Balance $O$: There is a total of 5 oxygens on the reactans side. And 2 on $2KOH$, so, $O_2$ should add 3 to balance the equation: $H_2O + 2KO_2 -- \gt 2KOH + \frac{3}{2}O_2$ Multiply all the coefficients by 2, to remove the fraction: $2H_2O + 4KO_2 -- \gt 4KOH + 3O_2$ - The equation is balanced (e) 1. Start by balancing the kind of atoms that appear only one time in each side: $S$, $O$ and $N$: - Balance $N$: We have 2 nitrogens on the products side in $(NH_4)_2SO_4$, so, we should put that number as the coefficient of $NH_3$: $H_2SO_4 + 2NH_3 -- \gt (NH_4)_2SO_4$ - The equation is balanced