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What volume of carbon dioxide will 4.06 q of antacid made of calcium

g
carbonate produce at 37.0 °C and 1.00 atm in the stomach according
o the following reaction?
CaCO3 (s) + 2 HCI (aq) → CaCl₂ (aq) + H₂O (1) + CO₂ (g)

User Nop
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1 Answer

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The volume of \(CO_2\) produced at 37.0 °C and 1.00 atm is approximately 105.15 liters.

To calculate the volume of carbon dioxide (\(CO_2\)) produced, we can follow these steps:

1. **Find the moles of \(CO_2\) produced:**

\[ \text{moles of } CO_2 = \frac{\text{moles of CaCO}_3}{1} \]

\[ \text{moles of } CO_2 = \frac{4.06 \, \text{mol}}{1} \]

2. **Use the Ideal Gas Law to find the volume of \(CO_2\):**

\[ PV = nRT \]

\[ V = \frac{nRT}{P} \]

\[ V = \frac{(4.06 \, \text{mol})(0.0821 \, \text{L} \cdot \text{atm} / \text{mol} \cdot \text{K})(310.15 \, \text{K})}{1.00 \, \text{atm}} \]

3. **Solve for \(V\):**

\[ V = \frac{(4.06 \times 0.0821 \times 310.15)}{1.00} \]

\[ V \approx \frac{105.151}{1.00} \]

\[ V \approx 105.15 \, \text{L} \]

Therefore, the volume of \(CO_2\) produced at 37.0 °C and 1.00 atm is approximately 105.15 liters.

User Jon Shea
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