Q3.118CP

Question

Convert the following descriptions into balanced equations: (a) In a gaseous reaction, hydrogen sulfide burns in oxygen to form sulfur dioxide and water vapor. (b) When crystalline potassium chlorate is heated to just above its melting point, it reacts to form two different crystalline compounds, potassium chloride and potassium perchlorate. (c) When hydrogen gas is passed over powdered iron (III) oxide, iron metal and water vapor form. (d) The combustion of gaseous ethane in air forms carbon dioxide and water vapor. (e) Iron (II) chloride is converted to iron (III) fluoride by treatment with chlorine trifluoride gas. Chlorine gas is also formed.

Step-by-Step Solution

Verified

Answer

Normal equations can be converted into balanced equations.

1Step 1: Converting the equation into a balanced equation in a

a. Balance the number of O atoms:

$H_{2} S_{(g)} + \frac{3}{2} O_{2 (g)} \to S O_{2 (g)} + H_{2} O_{(g)} .$

On multiplying the whole equation by 2, we get:

$2 H_{2} S_{(g)} + 3 O_{2 (g)} \to 2 S O_{2 (g)} + 2 H_{2} O_{(g)} .$

2Step 2: Converting the equation into a balanced equation in b

On balancing the number of O atoms, we get:

$4 K C l O_{3 (s)} \to 3 K C l O_{4 (s)} + K C l_{(s)} .$

3Step 3: Converting the equation into a balanced equation in c

c. On balancing the number of O atoms, we get:

$H_{(2) g} + F e_{2} O_{3 (S)} \to 3 H_{2} O_{(g)} + F e_{(s)} .$

On balancing the number of H atoms, we get:

$3 H_{(2) g} + F e_{2} O_{3 (S)} \to 3 H_{2} O_{(g)} + F e_{(s)} .$

On balancing the number of Fe atoms, we get:

$3 H_{(2) g} + F e_{2} O_{3 (S)} \to 3 H_{2} O_{(g)} + 2 F e_{(s)} .$

4Step 4: Converting the equation into a balanced equation in d

d. On balancing the number of H, C, and O atoms one by one, we get:

$C_{2} H_{6 (g)} + O_{2 (g)} \to 3 H_{2} O_{(g)} + C O_{2 (g)} C_{2} H_{6 (g)} + O_{2 (g)} \to 3 H_{2} O_{(g)} + 2 C O_{2 (g)} C_{2} H_{6 (g)} + 7 / 2 O_{2 (g)} \to 3 H_{2} O_{(g)} + 2 C O_{2 (g)} 2 C_{2} H_{6 (g)} + 7 O_{2 (g)} \to 6 H_{2} O_{(g)} + 4 C O_{2 (g)} .$

5Step 5: Converting the equation into a balanced equation in e

e. On balancing the number of Cl, F, and Fe atoms one by one, we get:

$F e C l_{2 (s)} + 2 C l F_{3 (g)} \to F e F_{3 (s)} + 2 C l_{2 (g)} F e C l_{2 (s)} + 2 C l F_{3 (g)} \to 2 F e F_{3 (s)} + 2 C l_{2 (g)} 2 F e C l_{2 (s)} + 2 C l F_{3 (g)} \to 2 F e F_{3 (s)} + 3 C l_{2 (g)} .$

Q3.115CP

Q3.119CP

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