Consider the following reaction:
`KClO_(3)(s)overset(Delta)toKCl(s)+O_(2)(g)`, yield`=60%`
`Zn(s)+H_(2)SO_(4)(aq)toZnSO_(4)(aq)+H_(2)(g)`, yield `=50%`
`2H_(2)(g)+O_(2)(g)to2H_(2)O(l)`, yield`=50%`
What volume of 0.2 M `H_(2)SO_(4)` solution is required to produce enough `H_(2)` to completely react with `O_(2)` liberated due decomposition of `1.225 g KClO_(3)`.
(Molecular weight of `KClO_(3)=39+53.5+3xx16`
`=122.5g mol^(-1)`)

Options

A. 150 mL
B. 180 mL
C. 360 mL
D. 480 mL

Explanation

To solve the problem step by step, we need to follow the stoichiometric relationships in the given chemical reactions and the yields provided.

### Step 1: Calculate the number of moles of KClO3
Given the mass of KClO3 = 1.225 g and its molecular weight = 122.5 g/mol.

\[ \text{Number of moles of KClO3} = \frac{\text{mass}}{\text{molecular weight}} = \frac{1.225 \, \text{g}}{122.5 \, \text{g/mol}} = 0.01 \, \text{mol} \]

### Step 2: Determine the amount of O2 produced from KClO3
From the balanced reaction:
\[ 2 \, KClO_3 (s) \rightarrow 2 \, KCl (s) + 3 \, O_2 (g) \]
This means that 2 moles of KClO3 produce 3 moles of O2. Therefore, 1 mole of KClO3 will produce:
\[ \frac{3}{2} \, \text{moles of O2} \]
Thus, for 0.01 moles of KClO3:
\[ \text{Moles of O2} = 0.01 \, \text{mol} \times \frac{3}{2} = 0.015 \, \text{mol} \]

### Step 3: Adjust for the yield of the reaction
The yield of the reaction is given as 60%. Therefore, the actual moles of O2 produced will be:
\[ \text{Actual moles of O2} = 0.015 \, \text{mol} \times \frac{60}{100} = 0.009 \, \text{mol} \]

### Step 4: Calculate the moles of H2 required to react with O2
From the reaction:
\[ 2 \, H_2 (g) + O_2 (g) \rightarrow 2 \, H_2O (l) \]
The stoichiometry shows that 1 mole of O2 reacts with 2 moles of H2. Therefore, for 0.009 moles of O2:
\[ \text{Moles of H2 required} = 0.009 \, \text{mol} \times 2 = 0.018 \, \text{mol} \]

### Step 5: Calculate the moles of H2SO4 required
From the reaction:
\[ Zn (s) + H_2SO_4 (aq) \rightarrow ZnSO_4 (aq) + H_2 (g) \]
The stoichiometry shows that 1 mole of H2SO4 produces 1 mole of H2. Therefore, to produce 0.018 moles of H2, we need:
\[ \text{Moles of H2SO4 required} = 0.018 \, \text{mol} \]

### Step 6: Adjust for the yield of H2SO4 reaction
The yield of this reaction is given as 50%. Therefore, to produce 0.018 moles of H2, we need:
\[ \text{Moles of H2SO4 required} = 0.018 \, \text{mol} \times \frac{100}{50} = 0.036 \, \text{mol} \]

### Step 7: Convert moles of H2SO4 to volume
The concentration of H2SO4 solution is given as 0.2 M. We can use the formula:
\[ \text{Volume (L)} = \frac{\text{moles}}{\text{concentration}} = \frac{0.036 \, \text{mol}}{0.2 \, \text{mol/L}} = 0.18 \, \text{L} \]
Converting to milliliters:
\[ \text{Volume (mL)} = 0.18 \, \text{L} \times 1000 \, \text{mL/L} = 180 \, \text{mL} \]

### Final Answer:
The volume of 0.2 M H2SO4 solution required is **180 mL**.

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