balancing equations ws #4
name:
block:
balance and identify the following equations.
1. ___ $kclo_3 \
ightarrow$ ___ $kcl +$ ___ $o_2$

2. ___ $albr_3 +$ ___ $k_2so_4 \
ightarrow$ ___ $kbr +$ ___ $al_2(so_4)_3$

3. ___ $ch_4 +$ ___ $o_2 \
ightarrow$___ $co_2 +$ ___$h_2o$

4. ___ $fe +$ ___ $o_2 \
ightarrow$ ___ $fe_2o_3$

5. ___ $al +$ ___ $feo \
ightarrow$ ___ $al_2o_3 +$ ___ $fe$

6. ___$fe_2(so_4)_3 +$ ___$koh \
ightarrow$ ___ $k_2so_4 +$ ___ $fe(oh)_3$

7. ___$p_4 +$ ___ $o_2 \
ightarrow$___ $p_2o_5$

8. ___ $k +$ ___ $br_2 \
ightarrow$ ___ $kbr$

9. ___ $s_8 +$ ___ $o_2 \
ightarrow$ ___ $so_3$

10. ___ $c_7h_{16} +$ ___ $o_2 \
ightarrow$ ___ $co_2 +$ ___$h_2o$

11. ____ $fecl_3 +$ ____ $naoh \
ightarrow$ ____ $fe(oh)_3 +$ ____$nacl$

12. ____ $k +$ ____ $mgbr_2 \
ightarrow$ ____ $kbr +$ ____ $mg$

Question

balancing equations ws #4
name:
block:
balance and identify the following equations.
1. ___ $kclo_3 \
ightarrow$ ___ $kcl +$ ___ $o_2$

2. ___ $albr_3 +$ ___ $k_2so_4 \
ightarrow$ ___ $kbr +$ ___ $al_2(so_4)_3$

3. ___ $ch_4 +$ ___ $o_2 \
ightarrow$___ $co_2 +$ ___$h_2o$

4. ___ $fe +$ ___ $o_2 \
ightarrow$ ___ $fe_2o_3$

5. ___ $al +$ ___ $feo \
ightarrow$ ___ $al_2o_3 +$ ___ $fe$

6. ___$fe_2(so_4)_3 +$ ___$koh \
ightarrow$ ___ $k_2so_4 +$ ___ $fe(oh)_3$

7. ___$p_4 +$ ___ $o_2 \
ightarrow$___ $p_2o_5$

8. ___ $k +$ ___ $br_2 \
ightarrow$ ___ $kbr$

9. ___ $s_8 +$ ___ $o_2 \
ightarrow$ ___ $so_3$

10. ___ $c_7h_{16} +$ ___ $o_2 \
ightarrow$ ___ $co_2 +$ ___$h_2o$

11. ____ $fecl_3 +$ ____ $naoh \
ightarrow$ ____ $fe(oh)_3 +$ ____$nacl$

12. ____ $k +$ ____ $mgbr_2 \
ightarrow$ ____ $kbr +$ ____ $mg$

Accepted Answer

### Problem 1: $\boldsymbol{\ce{KClO3 -> KCl + O2}}$
# Explanation:
## Step1: Balance K and Cl (already 1 each)
Start with K and Cl: 1 $\ce{KClO3}$ has 1 K and 1 Cl, same as $\ce{KCl}$.
## Step2: Balance O. $\ce{KClO3}$ has 3 O, $\ce{O2}$ is diatomic. Let coefficient of $\ce{KClO3}$ be 2, then O: 2×3 = 6. So $\ce{O2}$ needs 3 (3×2=6). Then KCl: 2 (from 2 $\ce{KClO3}$).
$2\ce{KClO3}
ightarrow 2\ce{KCl} + 3\ce{O2}$
# Answer:
2, 2, 3

### Problem 2: $\boldsymbol{\ce{AlBr3 + K2SO4 -> KBr + Al2(SO4)3}}$
# Explanation:
## Step1: Balance Al. $\ce{Al2(SO4)3}$ has 2 Al, so $\ce{AlBr3}$ needs 2.
## Step2: Balance $\ce{SO4^{2-}}$. $\ce{Al2(SO4)3}$ has 3, so $\ce{K2SO4}$ needs 3.
## Step3: Balance K. 3 $\ce{K2SO4}$ has 6 K, so $\ce{KBr}$ needs 6. Then Br: 2 $\ce{AlBr3}$ has 6 Br, which matches 6 $\ce{KBr}$.
$2\ce{AlBr3} + 3\ce{K2SO4}
ightarrow 6\ce{KBr} + 1\ce{Al2(SO4)3}$
# Answer:
2, 3, 6, 1

### Problem 3: $\boldsymbol{\ce{CH4 + O2 -> CO2 + H2O}}$
# Explanation:
## Step1: Balance C (1 each). Balance H: $\ce{CH4}$ has 4 H, so $\ce{H2O}$ needs 2 (2×2=4).
## Step2: Balance O. $\ce{CO2}$ has 2, $\ce{H2O}$ (2) has 2, total O: 4. So $\ce{O2}$ needs 2 (2×2=4).
$1\ce{CH4} + 2\ce{O2}
ightarrow 1\ce{CO2} + 2\ce{H2O}$
# Answer:
1, 2, 1, 2

### Problem 4: $\boldsymbol{\ce{Fe + O2 -> Fe2O3}}$
# Explanation:
## Step1: Balance Fe. $\ce{Fe2O3}$ has 2 Fe, so Fe needs 2.
## Step2: Balance O. $\ce{Fe2O3}$ has 3 O, $\ce{O2}$ is diatomic. Let coefficient of $\ce{Fe2O3}$ be 2, then O: 6. So $\ce{O2}$ needs 3 (3×2=6). Then Fe: 4 (2×2).
$4\ce{Fe} + 3\ce{O2}
ightarrow 2\ce{Fe2O3}$
# Answer:
4, 3, 2

### Problem 5: $\boldsymbol{\ce{Al + FeO -> Al2O3 + Fe}}$
# Explanation:
## Step1: Balance Al. $\ce{Al2O3}$ has 2 Al, so Al needs 2.
## Step2: Balance O. $\ce{Al2O3}$ has 3 O, so $\ce{FeO}$ needs 3 (3×1=3).
## Step3: Balance Fe. 3 $\ce{FeO}$ has 3 Fe, so Fe needs 3.
$2\ce{Al} + 3\ce{FeO}
ightarrow 1\ce{Al2O3} + 3\ce{Fe}$
# Answer:
2, 3, 1, 3

### Problem 6: $\boldsymbol{\ce{Fe2(SO4)3 + KOH -> K2SO4 + Fe(OH)3}}$
# Explanation:
## Step1: Balance Fe. $\ce{Fe2(SO4)3}$ has 2 Fe, so $\ce{Fe(OH)3}$ needs 2.
## Step2: Balance $\ce{SO4^{2-}}$. $\ce{Fe2(SO4)3}$ has 3, so $\ce{K2SO4}$ needs 3.
## Step3: Balance K. 3 $\ce{K2SO4}$ has 6 K, so $\ce{KOH}$ needs 6. Then OH: 6 $\ce{KOH}$ has 6 OH, which matches 2 $\ce{Fe(OH)3}$ (2×3=6).
$1\ce{Fe2(SO4)3} + 6\ce{KOH}
ightarrow 3\ce{K2SO4} + 2\ce{Fe(OH)3}$
# Answer:
1, 6, 3, 2

### Problem 7: $\boldsymbol{\ce{P4 + O2 -> P2O5}}$
# Explanation:
## Step1: Balance P. $\ce{P4}$ has 4 P, $\ce{P2O5}$ has 2 P, so $\ce{P2O5}$ needs 2 (2×2=4).
## Step2: Balance O. $\ce{P2O5}$ (2) has 10 O, so $\ce{O2}$ needs 5 (5×2=10).
$1\ce{P4} + 5\ce{O2}
ightarrow 2\ce{P2O5}$
# Answer:
1, 5, 2

### Problem 8: $\boldsymbol{\ce{K + Br2 -> KBr}}$
# Explanation:
## Step1: Balance Br. $\ce{Br2}$ has 2 Br, so $\ce{KBr}$ needs 2.
## Step2: Balance K. 2 $\ce{KBr}$ has 2 K, so K needs 2.
$2\ce{K} + 1\ce{Br2}
ightarrow 2\ce{KBr}$
# Answer:
2, 1, 2

### Problem 9: $\boldsymbol{\ce{S8 + O2 -> SO3}}$
# Explanation:
## Step1: Balance S. $\ce{S8}$ has 8 S, so $\ce{SO3}$ needs 8.
## Step2: Balance O. $\ce{SO3}$ (8) has 24 O, so $\ce{O2}$ needs 12 (12×2=24).
$1\ce{S8} + 12\ce{O2}
ightarrow 8\ce{SO3}$
# Answer:
1, 12, 8

### Problem 10: $\boldsymbol{\ce{C7H16 + O2 -> CO2 + H2O}}$
# Explanation:
## Step1: Balance C. $\ce{C7H16}$ has 7 C, so $\ce{CO2}$ needs 7.
## Step2: Balance H. $\ce{C7H16}$ has 16 H, so $\ce{H2O}$ needs 8 (8×2=16).
## Step3: Balance O. $\ce{CO2}$ (7) has 14, $\ce{H2O}$ (8) has 8, total O: 22. So $\ce{O2}$ needs 11 (11×2=22).
$1\ce{C7H16} + 11\ce{O2}
ightarrow 7\ce{CO2} + 8\ce{H2O}$
# Answer:
1, 11, 7, 8

### Problem 11: $\boldsymbol{\ce{FeCl3 + NaOH -> Fe(OH)3 + NaCl}}$
# Explanation:
## Step1: Balance Fe (1 each). Balance Cl: $\ce{FeCl3}$ has 3 Cl, so $\ce{NaCl}$ needs 3.
## Step2: Balance Na. 3 $\ce{NaCl}$ has 3 Na, so $\ce{NaOH}$ needs 3. Then OH: 3 $\ce{NaOH}$ has 3 OH, which matches $\ce{Fe(OH)3}$ (1×3=3).
$1\ce{FeCl3} + 3\ce{NaOH}
ightarrow 1\ce{Fe(OH)3} + 3\ce{NaCl}$
# Answer:
1, 3, 1, 3

### Problem 12: $\boldsymbol{\ce{K + MgBr2 -> KBr + Mg}}$
# Explanation:
## Step1: Balance Br. $\ce{MgBr2}$ has 2 Br, so $\ce{KBr}$ needs 2.
## Step2: Balance K. 2 $\ce{KBr}$ has 2 K, so K needs 2. Then Mg: 1 (from 1 $\ce{MgBr2}$) matches 1 Mg.
$2\ce{K} + 1\ce{MgBr2}
ightarrow 2\ce{KBr} + 1\ce{Mg}$
# Answer:
2, 1, 2, 1

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QUESTION IMAGE

Question

Explanation:

Problem 1: $\boldsymbol{\ce{KClO3 -> KCl + O2}}$

Step1: Balance K and Cl (already 1 each)

Step2: Balance O. $\ce{KClO3}$ has 3 O, $\ce{O2}$ is diatomic. Let coefficient of $\ce{KClO3}$ be 2, then O: 2×3 = 6. So $\ce{O2}$ needs 3 (3×2=6). Then KCl: 2 (from 2 $\ce{KClO3}$).

Step1: Balance Al. $\ce{Al2(SO4)3}$ has 2 Al, so $\ce{AlBr3}$ needs 2.

Step2: Balance $\ce{SO4^{2-}}$. $\ce{Al2(SO4)3}$ has 3, so $\ce{K2SO4}$ needs 3.

Step3: Balance K. 3 $\ce{K2SO4}$ has 6 K, so $\ce{KBr}$ needs 6. Then Br: 2 $\ce{AlBr3}$ has 6 Br, which matches 6 $\ce{KBr}$.

Step1: Balance C (1 each). Balance H: $\ce{CH4}$ has 4 H, so $\ce{H2O}$ needs 2 (2×2=4).

Step2: Balance O. $\ce{CO2}$ has 2, $\ce{H2O}$ (2) has 2, total O: 4. So $\ce{O2}$ needs 2 (2×2=4).

Answer:

Problem 2: $\boldsymbol{\ce{AlBr3 + K2SO4 -> KBr + Al2(SO4)3}}$