Question

part c: balance each of the following equations following the procedure described in class. be sure to show your work.
p + o₂ → p₄o₁₀
p=
o=
mg + o₂ → mgo
mg=
o=
hgo → hg + o₂
hg=
o=
al₂o₃ → al + o₂
al=
o=
bacl₂ + h₂so₄ → baso₄ + hcl
ba=
cl=
h=
s=
o=
ba=
cl=
h=
s=
o=

Explanation:

Step1: Balance P in $P + O_2

ightarrow P_4O_{10}$
There are 4 P atoms in $P_4O_{10}$, so we put 4 in front of P: $4P + O_2
ightarrow P_4O_{10}$

Step2: Balance O in $4P + O_2

ightarrow P_4O_{10}$
There are 10 O atoms in $P_4O_{10}$ and 2 in $O_2$. We put 5 in front of $O_2$: $4P+5O_2 = P_4O_{10}$

Step3: Balance Mg in $Mg + O_2

ightarrow MgO$
There are 2 O atoms in $O_2$ and 1 in MgO. First, put 2 in front of MgO: $Mg + O_2
ightarrow 2MgO$. Then put 2 in front of Mg: $2Mg+O_2 = 2MgO$

Step4: Balance O in $HgO

ightarrow Hg + O_2$
There are 2 O atoms in $O_2$ and 1 in HgO. Put 2 in front of HgO: $2HgO
ightarrow Hg + O_2$. Then put 2 in front of Hg: $2HgO=2Hg + O_2$

Step5: Balance O in $Al_2O_3

ightarrow Al + O_2$
The least - common multiple of 3 and 2 (number of O atoms) is 6. Put 2 in front of $Al_2O_3$ and 3 in front of $O_2$: $2Al_2O_3
ightarrow Al + 3O_2$. Then put 4 in front of Al: $2Al_2O_3 = 4Al+3O_2$

Step6: Balance $BaCl_2 + H_2SO_4

ightarrow BaSO_4 + HCl$
The number of Ba, S and O atoms is already balanced. There are 2 Cl atoms in $BaCl_2$ and 1 in HCl, and 2 H atoms in $H_2SO_4$ and 1 in HCl. Put 2 in front of HCl: $BaCl_2 + H_2SO_4=BaSO_4 + 2HCl$

Answer:

$4P + 5O_2=P_4O_{10}$
$2Mg+O_2 = 2MgO$
$2HgO=2Hg + O_2$
$2Al_2O_3 = 4Al+3O_2$
$BaCl_2 + H_2SO_4=BaSO_4 + 2HCl$