Question

4.
identifying reaction types & balancing equations
step 2: balance the following chemical equations:
n₂ + h₂ → __ nh₃
kclo₃ → kcl + __ o₂
nacl + f₂ → naf + cl₂
h₂ + o₂ → __ h₂o
agno₃ + mgcl₂ → agcl + mg(no₃)₂
albr₃ + k₂so₄ → kbr + al₂(so₄)₃
ch₄ + o₂ → co₂ + h₂o
c₃h₈ + o₂ → co₂ + h₂o
fecl₃ + naoh → fe(oh)₃ + nacl
p + o₂ → __ p₂o₅
na + h₂o → naoh + h₂
ag₂o → ag + __ o₂
co₂ + h₂o → c₆h₁₂o₆ + o₂
k + mgbr₂ → kbr + mg
hno₃(aq) + ba(oh)₂(aq) → ba(no₃)₂(aq) + h₂o (l)
c₅h₁₂(g) + o₂(g) → co₂(g) + h₂o(g)
al(s) + fe₂o₃(aq) → al₂o₃(aq) + fe(s)
al(s) + o₂(g) → __ al₂o₃(s)

Explanation:

Step 1: Balance $\boldsymbol{N_2 + H_2

ightarrow NH_3}$
- Count N: Left has 2, right has 1. So put 2 in front of $NH_3$. Now N: 2 on both.
- Count H: Right has $2\times3 = 6$, left has 2. So put 3 in front of $H_2$.
Balanced: $1\ N_2 + 3\ H_2
ightarrow 2\ NH_3$

Step 2: Balance $\boldsymbol{KClO_3

ightarrow KCl + O_2}$
- Count K: 1 on left, 1 on right. Good.
- Count Cl: 1 on left, 1 on right. Good.
- Count O: Left has 3, right has 2. Find LCM of 3 and 2 (6). Put 2 in front of $KClO_3$ (O: 6) and 3 in front of $O_2$ (O: 6). Then K and Cl: put 2 in front of $KCl$.
Balanced: $2\ KClO_3
ightarrow 2\ KCl + 3\ O_2$

Step 3: Balance $\boldsymbol{NaCl + F_2

ightarrow NaF + Cl_2}$
- Count Na: 1 on left, 1 on right.
- Count Cl: Left has 1, right has 2. Put 2 in front of $NaCl$. Now Cl: 2 on left.
- Count F: Right has 1, left has 2. Put 2 in front of $NaF$. Now F: 2 on right.
Balanced: $2\ NaCl + 1\ F_2
ightarrow 2\ NaF + 1\ Cl_2$

Step 4: Balance $\boldsymbol{H_2 + O_2

ightarrow H_2O}$
- Count H: 2 on left, 2 on right (per $H_2O$).
- Count O: Left has 2, right has 1. Put 2 in front of $H_2O$ (O: 2). Now H: $2\times2 = 4$, so put 2 in front of $H_2$.
Balanced: $2\ H_2 + 1\ O_2
ightarrow 2\ H_2O$

Step 5: Balance $\boldsymbol{AgNO_3 + MgCl_2

ightarrow AgCl + Mg(NO_3)_2}$
- Count Ag: 1 on left, 1 on right.
- Count Mg: 1 on left, 1 on right.
- Count $NO_3$: Right has 2, left has 1. Put 2 in front of $AgNO_3$. Now $NO_3$: 2 on left.
- Count Cl: Left has 2 (from $MgCl_2$), right has 1 (from $AgCl$). Put 2 in front of $AgCl$.
Balanced: $2\ AgNO_3 + 1\ MgCl_2
ightarrow 2\ AgCl + 1\ Mg(NO_3)_2$

Step 6: Balance $\boldsymbol{AlBr_3 + K_2SO_4

ightarrow KBr + Al_2(SO_4)_3}$
- Count Al: Left has 1, right has 2. Put 2 in front of $AlBr_3$. Now Al: 2 on left.
- Count $SO_4$: Right has 3, left has 1. Put 3 in front of $K_2SO_4$. Now $SO_4$: 3 on left.
- Count K: Left has $3\times2 = 6$, right has 1. Put 6 in front of $KBr$.
Balanced: $2\ AlBr_3 + 3\ K_2SO_4
ightarrow 6\ KBr + 1\ Al_2(SO_4)_3$

Step 7: Balance $\boldsymbol{CH_4 + O_2

ightarrow CO_2 + H_2O}$
- Count C: 1 on left, 1 on right. Good.
- Count H: Left has 4, right has 2 (per $H_2O$). Put 2 in front of $H_2O$. Now H: 4 on right.
- Count O: Right has $2 + 2\times1 = 4$, left has 2. Put 2 in front of $O_2$.
Balanced: $1\ CH_4 + 2\ O_2
ightarrow 1\ CO_2 + 2\ H_2O$

Step 8: Balance $\boldsymbol{C_3H_8 + O_2

ightarrow CO_2 + H_2O}$
- Count C: Left has 3, right has 1. Put 3 in front of $CO_2$.
- Count H: Left has 8, right has 2. Put 4 in front of $H_2O$ (H: 8).
- Count O: Right has $3\times2 + 4\times1 = 10$, left has 2. Put 5 in front of $O_2$ (O: 10).
Balanced: $1\ C_3H_8 + 5\ O_2
ightarrow 3\ CO_2 + 4\ H_2O$

Step 9: Balance $\boldsymbol{FeCl_3 + NaOH

ightarrow Fe(OH)_3 + NaCl}$
- Count Fe: 1 on left, 1 on right. Good.
- Count Cl: Left has 3, right has 1. Put 3 in front of $NaCl$. Now Cl: 3 on right.
- Count Na: Right has 3, left has 1. Put 3 in front of $NaOH$. Now Na: 3 on left.
- Count OH: Left has 3 (from $3\ NaOH$), right has 3 (from $Fe(OH)_3$). Good.
Balanced: $1\ FeCl_3 + 3\ NaOH
ightarrow 1\ Fe(OH)_3 + 3\ NaCl$

Step 10: Balance $\boldsymbol{P + O_2

ightarrow P_2O_5}$
- Count P: Left has 1, right has 2. Put 2 in front of $P$. Now P: 2 on left.
- Count O: Right has 5, left has 2. Find LCM of 5 and 2 (10). Put 5 in front of $O_2$ (O: 10) and 2 in front of $P_2O_5$ (O: 10). Now P: $2\times2 = 4$, so put 4 in front of $P$.
Balanced: $4\ P + 5\ O_2
ightarrow 2\ P_2O_5$

Step 11: Balance $\boldsymbol{Na + H_2O

ightarrow NaOH + H_2}$
- Count Na: 1 on left, 1 on right.
- Count O: 1 on left, 1 on right.
- Count H: Left has 2, right has $1 + 2 = 3$. Put 2 in front of $NaOH$ (H: 2 + 2 = 4) and 2 in front of $H_2O$ (H: 4). Now Na: put 2 in front of $Na$.
Balanced: $2\ Na + 2\ H_2O
ightarrow 2\ NaOH + 1\ H_2$

Step 12: Balance $\boldsymbol{Ag_2O

ightarrow Ag + O_2}$
- Count Ag: Left has 2, right has 1. Put 2 in front of $Ag$. Now Ag: 2 on right.
- Count O: Left has 1, right has 2. Put 2 in front of $Ag_2O$ (O: 2) and 1 in front of $O_2$ (O: 2). Now Ag: $2\times2 = 4$, so put 4 in front of $Ag$.
Balanced: $2\ Ag_2O
ightarrow 4\ Ag + 1\ O_2$

Step 13: Balance $\boldsymbol{CO_2 + H_2O

ightarrow C_6H_{12}O_6 + O_2}$
- Count C: Left has 1, right has 6. Put 6 in front of $CO_2$. Now C: 6 on left.
- Count H: Left has 2, right has 12. Put 6 in front of $H_2O$ (H: 12). Now H: 12 on left.
- Count O: Left has $6\times2 + 6\times1 = 18$, right has $6 + 2x$ (x = $O_2$ coefficient). $6 + 2x = 18 \Rightarrow x = 6$.
Balanced: $6\ CO_2 + 6\ H_2O
ightarrow 1\ C_6H_{12}O_6 + 6\ O_2$

Step 14: Balance $\boldsymbol{K + MgBr_2

ightarrow KBr + Mg}$
- Count K: 1 on left, 1 on right.
- Count Mg: 1 on left, 1 on right.
- Count Br: Left has 2, right has 1. Put 2 in front of $KBr$. Now Br: 2 on right. Then put 2 in front of $K$.
Balanced: $2\ K + 1\ MgBr_2
ightarrow 2\ KBr + 1\ Mg$

Step 15: Balance $\boldsymbol{HNO_3 + Ba(OH)_2

ightarrow Ba(NO_3)_2 + H_2O}$
- Count Ba: 1 on left, 1 on right. Good.
- Count $NO_3$: Right has 2, left has 1. Put 2 in front of $HNO_3$. Now $NO_3$: 2 on left.
- Count H: Left has $2 + 2 = 4$ (from $2\ HNO_3$ and $Ba(OH)_2$), right has 2 (per $H_2O$). Put 2 in front of $H_2O$ (H: 4).
Balanced: $2\ HNO_3 + 1\ Ba(OH)_2
ightarrow 1\ Ba(NO_3)_2 + 2\ H_2O$

Step 16: Balance $\boldsymbol{C_5H_{12} + O_2

ightarrow CO_2 + H_2O}$
- Count C: Left has 5, right has 1. Put 5 in front of $CO_2$.
- Count H: Left has 12, right has 2. Put 6 in front of $H_2O$ (H: 12).
- Count O: Right has $5\times2 + 6\times1 = 16$, left has 2. Put 8 in front of $O_2$ (O: 16).
Balanced: $1\ C_5H_{12} + 8\ O_2
ightarrow 5\ CO_2 + 6\ H_2O$

Step 17: Balance $\boldsymbol{Al + Fe_2O_3

ightarrow Al_2O_3 + Fe}$
- Count Al: Left has 1, right has 2. Put 2 in front of $Al$. Now Al: 2 on left.
- Count Fe: Left has 2, right has 1. Put 2 in front of $Fe$. Now Fe: 2 on right.
Balanced: $2\ Al + 1\ Fe_2O_3
ightarrow 1\ Al_2O_3 + 2\ Fe$

Step 18: Balance $\boldsymbol{Al + O_2

ightarrow Al_2O_3}$
- Count Al: Left has 1, right has 2. Put 2 in front of $Al$. Now Al: 2 on left.
- Count O: Right has 3, left has 2. LCM of 3 and 2 is 6. Put 3 in front of $O_2$ (O: 6) and 2 in front of $Al_2O_3$ (O: 6). Now Al: $2\times2 = 4$, so put 4 in front of $Al$.
Balanced: $4\ Al + 3\ O_2
ightarrow 2\ Al_2O_3$

Answers (Coefficients in order):

1. $1, 3, 2$
2. $2, 2, 3$
3. $2, 1, 2, 1$
4. $2, 1, 2$
5. $2, 1, 2, 1$
6. $2, 3, 6, 1$
7. $1, 2, 1, 2$
8. $1, 5, 3, 4$
9. $1, 3, 1, 3$
10. $4, 5, 2$
11. $2, 2, 2, 1$
12. $2, 4, 1$
13. $6, 6, 1, 6$
14. $2, 1, 2, 1$
15. $2, 1, 1, 2$
16. $1, 8, 5, 6$
17. $2, 1, 1, 2$
18. $4, 3, 2$

Answer:

s (Coefficients in order):

1. $1, 3, 2$
2. $2, 2, 3$
3. $2, 1, 2, 1$
4. $2, 1, 2$
5. $2, 1, 2, 1$
6. $2, 3, 6, 1$
7. $1, 2, 1, 2$
8. $1, 5, 3, 4$
9. $1, 3, 1, 3$
10. $4, 5, 2$
11. $2, 2, 2, 1$
12. $2, 4, 1$
13. $6, 6, 1, 6$
14. $2, 1, 2, 1$
15. $2, 1, 1, 2$
16. $1, 8, 5, 6$
17. $2, 1, 1, 2$
18. $4, 3, 2$