Maak steeds een kloppende reactievergelijking door de juiste coëfficiënten in te vullen. a $$ \ldots \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\ldots \mathrm{F}_{2}(\mathrm{~g}) \rightarrow \ldots \mathrm{HF}(\mathrm{aq})+\ldots \mathrm{O}_{2}(\mathrm{~g}) $$ b $$ \ldots \mathrm{NH}_{3}(\mathrm{~g})+\ldots \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \ldots \mathrm{NO}(\mathrm{g})+\ldots \mathrm{H}_{2} \mathrm{O}(\mathrm{II}) $$ c $$ \ldots \mathrm{CuO}(\mathrm{s})+\ldots \mathrm{CH}_{4}(\mathrm{~g}) \rightarrow \ldots \mathrm{Cu}(\mathrm{s})+\ldots \mathrm{CO}_{2}(\mathrm{~g})+\ldots \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) $$ d $$ \ldots \mathrm{FeS}(\mathrm{s})+\ldots \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \ldots \mathrm{SO}_{2}(\mathrm{~g})+\ldots \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{~s}) $$

Question

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a) We write the unbalanced equation as  
$$
a\,\mathrm{H}_2\mathrm{O}(l) + b\,\mathrm{F}_2(g) \rightarrow c\,\mathrm{HF}(aq) + d\,\mathrm{O}_2(g).
$$

**Balancing step by step:**

1. The hydrogen atoms: water has 2 H, and HF has 1 H. Thus,
   $$
   2a = c.
   $$
2. The fluorine atoms: $$\mathrm{F}_2$$ supplies 2 F per molecule, and HF has 1 F. Thus,
   $$
   2b = c.
   $$
   This tells us $$2a = 2b$$ so $$a = b$$.

3. The oxygen atoms: water supplies $$a$$ oxygen atoms and $$\mathrm{O}_2$$ provides 2 per molecule, so:
   $$
   a = 2d.
   $$

Choosing $$a = 2$$ (the smallest integer choice), then $$b = 2$$, $$d = 1$$ and from (1) $$c = 2a = 4$$.

The balanced equation is  
$$
2\,\mathrm{H}_2\mathrm{O}(l) + 2\,\mathrm{F}_2(g) \rightarrow 4\,\mathrm{HF}(aq) + \mathrm{O}_2(g).
$$

---

b) Write the unbalanced equation as  
$$
a\,\mathrm{NH}_3(g) + b\,\mathrm{O}_2(g) \rightarrow c\,\mathrm{NO}(g) + d\,\mathrm{H}_2\mathrm{O}(l).
$$

**Balancing step by step:**

1. Nitrogen: ammonia and NO both contain 1 N. Thus,
   $$
   a = c.
   $$
2. Hydrogen: $$\mathrm{NH}_3$$ has 3 H and water has 2 H:
   $$
   3a = 2d \quad\Longrightarrow\quad d = \frac{3}{2}a.
   $$
3. Oxygen: $$\mathrm{O}_2$$ supplies 2 O per molecule. In the products, NO has 1 O each and water has 1 O each:
   $$
   2b = c + d.
   $$
   Substitute $$c = a$$ and $$d = \frac{3}{2}a$$:
   $$
   2b = a + \frac{3}{2}a = \frac{5}{2}a \quad\Longrightarrow\quad b = \frac{5}{4}a.
   $$

To have integer coefficients, choose $$a = 4$$. Then $$c = 4$$, $$d = 6$$ and $$b = 5$$.

The balanced equation is  
$$
4\,\mathrm{NH}_3(g) + 5\,\mathrm{O}_2(g) \rightarrow 4\,\mathrm{NO}(g) + 6\,\mathrm{H}_2\mathrm{O}(l).
$$

---

c) Write the unbalanced equation as  
$$
a\,\mathrm{CuO}(s) + b\,\mathrm{CH}_4(g) \rightarrow c\,\mathrm{Cu}(s) + d\,\mathrm{CO}_2(g) + e\,\mathrm{H}_2\mathrm{O}(l).
$$

**Balancing step by step:**

1. Copper: Only present in $$\mathrm{CuO}$$ and $$\mathrm{Cu}$$:
   $$
   a = c.
   $$
2. Carbon: Only in $$\mathrm{CH}_4$$ and $$\mathrm{CO}_2$$:
   $$
   b = d.
   $$
3. Hydrogen: $$\mathrm{CH}_4$$ provides 4 H and water provides 2 H:
   $$
   4b = 2e \quad\Longrightarrow\quad e = 2b.
   $$
4. Oxygen: In $$\mathrm{CuO}$$ there is 1 O per molecule and on the right $$\mathrm{CO}_2$$ provides 2 O per molecule and $$\mathrm{H}_2\mathrm{O}$$ provides 1:
   $$
   a = 2d + e.
   $$
   Substitute $$d = b$$ and $$e = 2b$$:
   $$
   a = 2b + 2b = 4b.
   $$
   Also, from step 1, $$c = a = 4b$$.

Choose $$b = 1$$. Then $$a = 4$$, $$c = 4$$, $$d = 1$$, $$e = 2$$.

The balanced equation is  
$$
4\,\mathrm{CuO}(s) + \mathrm{CH}_4(g) \rightarrow 4\,\mathrm{Cu}(s) + \mathrm{CO}_2(g) + 2\,\mathrm{H}_2\mathrm{O}(l).
$$

---

d) Write the unbalanced equation as  
$$
a\,\mathrm{FeS}(s) + b\,\mathrm{O}_2(g) \rightarrow c\,\mathrm{SO}_2(g) + d\,\mathrm{Fe}_2\mathrm{O}_3(s).
$$

**Balancing step by step:**

1. Iron: Only in $$\mathrm{FeS}$$ and $$\mathrm{Fe}_2\mathrm{O}_3$$:
   $$
   a = 2d.
   $$
2. Sulfur: Only in $$\mathrm{FeS}$$ and $$\mathrm{SO}_2$$:
   $$
   a = c.
   $$
3. Oxygen: On the left, $$\mathrm{O}_2$$ gives $$2b$$ atoms. On the right, each $$\mathrm{SO}_2$$ gives 2 O atoms and each $$\mathrm{Fe}_2\mathrm{O}_3$$ gives 3 O atoms:
   $$
   2b = 2c + 3d.
   $$
   Substitute $$c = a$$ and $$a = 2d$$:
   $$
   2b = 2(2d) + 3d = 4d + 3d = 7d \quad\Longrightarrow\quad b = \frac{7}{2}d.
   $$
To keep all coefficients integer, choose $$d = 2$$. Then $$a = 2d = 4$$, $$c = 4$$, and $$b = 7$$.

The balanced equation is  
$$
4\,\mathrm{FeS}(s) + 7\,\mathrm{O}_2(g) \rightarrow 4\,\mathrm{SO}_2(g) + 2\,\mathrm{Fe}_2\mathrm{O}_3(s).
$$
a) $$2\,\mathrm{H}_2\mathrm{O}(l) + 2\,\mathrm{F}_2(g) \rightarrow 4\,\mathrm{HF}(aq) + \mathrm{O}_2(g)$$ b) $$4\,\mathrm{NH}_3(g) + 5\,\mathrm{O}_2(g) \rightarrow 4\,\mathrm{NO}(g) + 6\,\mathrm{H}_2\mathrm{O}(l)$$ c) $$4\,\mathrm{CuO}(s) + \mathrm{CH}_4(g) \rightarrow 4\,\mathrm{Cu}(s) + \mathrm{CO}_2(g) + 2\,\mathrm{H}_2\mathrm{O}(l)$$ d) $$4\,\mathrm{FeS}(s) + 7\,\mathrm{O}_2(g) \rightarrow 4\,\mathrm{SO}_2(g) + 2\,\mathrm{Fe}_2\mathrm{O}_3(s)$$

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