The decomposition of cetian mass of `CaCO_(3)` gave `11.2dm^(3)` of `CO_(2)` gas at STP. The mass of KOH required to completely neutralise the gas is:

Options

A. 56g
B. 28g
C. 42g
D. 20g

Answer

To solve the problem of how much mass of KOH is required to completely neutralize the CO₂ gas produced from the decomposition of CaCO₃, we can follow these steps:

### Step 1: Calculate the number of moles of CO₂
Given the volume of CO₂ produced is 11.2 dm³ at STP (Standard Temperature and Pressure), we can use the fact that 1 mole of gas occupies 22.4 dm³ at STP.

\[ \text{Number of moles of CO₂} = \frac{\text{Volume of CO₂}}{\text{Molar volume at STP}} = \frac{11.2 \, \text{dm}^3}{22.4 \, \text{dm}^3/\text{mol}} = 0.5 \, \text{mol} \]

**Hint:** Remember that at STP, 1 mole of any gas occupies 22.4 liters.

### Step 2: Write the neutralization reaction
The reaction between KOH and CO₂ can be represented as follows:

\[ \text{KOH} + \text{CO₂} \rightarrow \text{KHCO₃} \]

From the balanced equation, we see that 1 mole of KOH reacts with 1 mole of CO₂.

### Step 3: Calculate the number of moles of KOH required
Since we have 0.5 moles of CO₂, we will need an equal number of moles of KOH to completely neutralize it.

\[ \text{Number of moles of KOH required} = 0.5 \, \text{mol} \]

**Hint:** The stoichiometry of the reaction shows a 1:1 ratio between KOH and CO₂.

### Step 4: Calculate the mass of KOH required
The molar mass of KOH is calculated as follows:

\[ \text{Molar mass of KOH} = \text{K} (39) + \text{O} (16) + \text{H} (1) = 56 \, \text{g/mol} \]

Now we can calculate the mass of KOH required using the formula:

\[ \text{Mass} = \text{Number of moles} \times \text{Molar mass} \]

Substituting the values:

\[ \text{Mass of KOH} = 0.5 \, \text{mol} \times 56 \, \text{g/mol} = 28 \, \text{g} \]

### Final Answer
The mass of KOH required to completely neutralize the CO₂ gas is **28 grams**.

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