90 g of acetic acid react with excess of `NaHCO_(3)` then what volume of `CO_(2)` will produce at S.T.P. Write your answer in terms of nearest integer.

Answer

To solve the problem of how much carbon dioxide (CO₂) is produced when 90 g of acetic acid (CH₃COOH) reacts with excess sodium bicarbonate (NaHCO₃), we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between acetic acid and sodium bicarbonate can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NaHCO}_3 \rightarrow \text{CH}_3\text{COONa} + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Calculate the molar mass of acetic acid
The molar mass of acetic acid (CH₃COOH) can be calculated as follows:
- Carbon (C): 12 g/mol × 2 = 24 g/mol
- Hydrogen (H): 1 g/mol × 4 = 4 g/mol
- Oxygen (O): 16 g/mol × 2 = 32 g/mol

Adding these together:
\[ 24 + 4 + 32 = 60 \text{ g/mol} \]

### Step 3: Calculate the number of moles of acetic acid
Using the mass of acetic acid given (90 g), we can find the number of moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{90 \text{ g}}{60 \text{ g/mol}} = 1.5 \text{ moles} \]

### Step 4: Determine the moles of CO₂ produced
From the balanced equation, we see that 1 mole of acetic acid produces 1 mole of carbon dioxide. Therefore, 1.5 moles of acetic acid will produce 1.5 moles of CO₂.

### Step 5: Calculate the volume of CO₂ at STP
At standard temperature and pressure (STP), 1 mole of any gas occupies 22.4 liters. Thus, the volume of CO₂ produced can be calculated as:
\[ \text{Volume of CO}_2 = \text{moles of CO}_2 \times 22.4 \text{ L/mol} \]
\[ \text{Volume of CO}_2 = 1.5 \text{ moles} \times 22.4 \text{ L/mol} = 33.6 \text{ liters} \]

### Step 6: Round to the nearest integer
The final answer, rounded to the nearest integer, is:
\[ \text{Volume of CO}_2 = 34 \text{ liters} \]

### Final Answer:
The volume of carbon dioxide produced at STP is **34 liters**.

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