Answer
a. Sn + 2NaOH $\longrightarrow$ $Na_{2}$Sn$O_{2}$+$H_{2}$
b. 8Al + 3$Fe_{3}$$O_{4}$ $\longrightarrow$ 4$Al_{2}$$O_{3}$+ 9Fe
c. 2C$H_{3}$OH+ 3$O_{2}$ $\longrightarrow$ 2C$O_{2}$ +4$H_{2}$O
d. $P_{4}$$O_{10}$+6 $H_{2}$O$\longrightarrow$ 4$H_{3}$P$O_{4}$
e. P$Cl_{5}$+4$H_{2}$O$\longrightarrow$ $H_{3}$P$O_{4}$ + 5HCl
Work Step by Step
The strategy requires to find the coefficient to balance the chemical equation. First look for atoms of elements that occur in only one substance on each side of equation.
Begin by balancing these atoms to get to the balanced chemical equation.
Count the atoms of each element in each side of equation, to see if they are in the same number in each side.
a. Sn + NaOH $\longrightarrow$ $Na_{2}$Sn$O_{2}$+$H_{2}$
1. Balance Na atoms by adding 2 before NaOH.
Sn + NaOH $\longrightarrow$ $Na_{2}$Sn$O_{2}$+$H_{2}$
Count the atoms of each elements and find our that we have balanced the equation.
b. Al + $Fe_{3}$$O_{4}$ $\longrightarrow$ $Al_{2}$$O_{3}$+ Fe
1. Start to balance Fe atoms by adding 3 before Fe.
Al + $Fe_{3}$$O_{4}$ $\longrightarrow$ $Al_{2}$$O_{3}$+ 3Fe
2. Balance O atoms as on the left are 4 and on the right side there are 3 atoms O. Add 3 before $Fe_{3}$$O_{4}$ ad 4 before $Al_{2}$$O_{3}$. This action will change the balance of Fe, so add 9 before Fe.
Al + 3$Fe_{3}$$O_{4}$ $\longrightarrow$ 4$Al_{2}$$O_{3}$+ 9Fe
4. Last step balance AL atoms as there is only 1 on the left, and 8 atoms on the right. Add 8 before Al.
Al + 3$Fe_{3}$$O_{4}$ $\longrightarrow$ 4$Al_{2}$$O_{3}$+ 9Fe
We have balanced the equation.
c. C$H_{3}$OH+ $O_{2}$ $\longrightarrow$ C$O_{2}$ +$H_{2}$O
1. Balance atoms of H as there are 4 on the left and 2 on the right. Add 2 before $H_{2}$O.
C$H_{3}$OH+ $O_{2}$ $\longrightarrow$ C$O_{2}$ +2$H_{2}$O
2- Balance O atoms as there are 3 on the left an 4 on the right. Add 2 before C$H_{3}$OH. As this would change the balance of C ( it would be 1 on the left, and 2 on the right, so add a 2 before C$O_{2}$ .
2C$H_{3}$OH+ $O_{2}$ $\longrightarrow$ 2C$O_{2}$+2$H_{2}$O
4. Re-balance H by adding 4 before $H_{2}$O and 3 before $O_{2}$
2C$H_{3}$OH+ 3$O_{2}$ $\longrightarrow$ 2C$O_{2}$+4$H_{2}$O
d. $P_{4}$$O_{10}$+ $H_{2}$O$\longrightarrow$ $H_{3}$P$O_{4}$
1. Balance P on the both sides by adding 4 before $H_{3}$P$O_{4}$
$P_{4}$$O_{10}$+ $H_{2}$O$\longrightarrow$ 4$H_{3}$P$O_{4}$
2. To balance H atoms add 6 before $H_{2}$O
$P_{4}$$O_{10}$+ $H_{2}$O$\longrightarrow$ 4$H_{3}$P$O_{4}$
We have balanced this chemical equation.
e. P$Cl_{5}$+$H_{2}$O$\longrightarrow$ $H_{3}$P$O_{4}$ + HCl
1. Balance Cl by adding 5 before HCl.
P$Cl_{5}$+$H_{2}$O$\longrightarrow$ $H_{3}$P$O_{4}$ + 5HCl
2. Balance H by adding 4 before $H_{2}$O.
P$Cl_{5}$+4$H_{2}$O$\longrightarrow$ $H_{3}$P$O_{4}$ + 5HCl
We have balanced this chemical equation.
