Answer
a. $O_{2}$ + 2P$Cl_{3}$ $\longrightarrow$ 2 PO$Cl_{3}$
b. $P_{4}$+ 6$N_{2}$O $\longrightarrow$ $P_{4}$$O_{6}$+6$N_{2}$
c. 2$As_{2}S_{3}$+9 $0_{2}$ $\longrightarrow$ 2$As_{2}$$O_{3}$+6$SO_{2}$
d. $Ca_{3}(PO_{4})_{2}$+ 4$H_{3}PO_{4}$ $\longrightarrow$
3$Ca(H_{2}PO_{4})_{2}$
Work Step by Step
The strategy for this exercises that requires to find the coefficient to balance the chemical equation is: Look for atoms of elements that occur in only one substance on each side of equation.
Begin by balancing these atoms to get to the balanced chemical equation.
Count the atoms of each element in each side of equation, to see if they are in the same number in each side.
a. We see that oxygen atoms are not balance as there are 2 on the left and only 1 on the right. So we add a 2 before PO$Cl_{3}$ .
$O_{2}$ + P$Cl_{3}$ $\longrightarrow$ 2 PO$Cl_{3}$
Now P and Cl atoms are not balance: P atoms is only 1 on the left and 2 on the right, while Cl there are 3 on the left and 2 x 3 = 6 on the right. To balance these atoms we need to add 2 before P$Cl_{3}$.
$O_{2}$ + 2P$Cl_{3}$ $\longrightarrow$ 2 PO$Cl_{3}$
Count the atoms of each element and find our that we have balanced the equation.
b- We see that oxygen atoms are not balance as there is 1 on the left and 6 on the right. So we add a 6 before $N_{2}$O.
$P_{4}$+ 6$N_{2}$O $\longrightarrow$ $P_{4}$$O_{6}$+$N_{2}$
Now we see that there are 6x2=12 atoms N on the left, and 2 on the right. So we add a 6 before $N_{2}$.
$P_{4}$+ 6$N_{2}$O $\longrightarrow$ $P_{4}$$O_{6}$+6$N_{2}$
Count the atoms of each element and find our that we have balanced the equation.
c. We see that S atoms are not balance as there are 3 on the left and 1 on the right. To balance this we add 3 before 3$SO_{2}$ :
$As_{2}S_{3}$+ $0_{2}$ $\longrightarrow$ $As_{2}$$O_{3}$+3$SO_{2}$
We see that O atoms are not balanced now, as we have only 2 on the left and 3+ 3x2= 9 on the right. We add 6 before $0_{2}$ and 2 before $As_{2}$$O_{3}$.
$As_{2}S_{3}$+6 $0_{2}$ $\longrightarrow$ 2$As_{2}$$O_{3}$+3$SO_{2}$
We now have 12 atoms of O on each side of equation, but we see that we changed the number of As atoms. To balance we have to add 2 before $As_{2}S_{3}$.
2$As_{2}S_{3}$+6 $0_{2}$ $\longrightarrow$ 2$As_{2}$$O_{3}$+3$SO_{2}$
Now we see that the balance of S atoms is not correct, so we change 3 to 6 before $SO_{2}$. This action will change the balance of O atoms, so we change 6 to 9 before $0_{2}$. So we will have:
2$As_{2}S_{3}$+9 $0_{2}$ $\longrightarrow$ 2$As_{2}$$O_{3}$+6$SO_{2}$
Count the atoms of each element and find our that we have balanced the equation.
d. We see that Ca atoms are not balance as there are 3 on the left and 1 on the right. Add 3 before $Ca(H_{2}PO_{4})_{2}$
$Ca_{3}(PO_{4})_{2}$+ $H_{3}PO_{4}$ $\longrightarrow$
3$Ca(H_{2}PO_{4})_{2}$
We see that H atoms are not balanced: there are 3 on the left and 2x2x3= 12 on the right. Add 4 before $H_{3}PO_{4}$ to balance:
$Ca_{3}(PO_{4})_{2}$+ 4$H_{3}PO_{4}$ $\longrightarrow$
3$Ca(H_{2}PO_{4})_{2}$
Count the atoms of each element and find our that we have balanced the equation.
