Question

sometimes in lab we collect the gas formed by a chemical reaction over water (see sketch at right). this makes it easy to isolate and measure the amount of gas produced. suppose the co₂ gas evolved by a certain chemical reaction taking place at 50.0 °c is collected over water, using an apparatus something like that in the sketch, and the final volume of gas in the collection tube is measured to be 98.6 ml. calculate the mass of co₂ that is in the collection tube. round your answer to 2 significant digits. you can make any normal and reasonable assumption about the reaction conditions and the nature of the gases. sketch of a gas - collection apparatus

Explanation:

Step1: Assume standard pressure (1 atm) and find moles of \( CO_2 \)

We use the ideal gas law \( PV = nRT \). First, convert volume to L: \( V = 98.6 \, \text{mL} = 0.0986 \, \text{L} \), temperature to Kelvin: \( T = 50.0 + 273.15 = 323.15 \, \text{K} \), \( R = 0.0821 \, \text{L·atm/(mol·K)} \), \( P = 1 \, \text{atm} \) (assuming standard pressure, and also assuming the vapor pressure of water is negligible or we consider dry \( CO_2 \) for simplicity, or we can account for water vapor, but let's proceed with dry \( CO_2 \) first).
\( n = \frac{PV}{RT} = \frac{1 \, \text{atm} \times 0.0986 \, \text{L}}{0.0821 \, \text{L·atm/(mol·K)} \times 323.15 \, \text{K}} \)
Calculate denominator: \( 0.0821 \times 323.15 \approx 26.53 \)
\( n \approx \frac{0.0986}{26.53} \approx 0.003716 \, \text{mol} \)

Step2: Calculate mass of \( CO_2 \)

Molar mass of \( CO_2 \) is \( 12.01 + 2 \times 16.00 = 44.01 \, \text{g/mol} \).
Mass \( m = n \times M = 0.003716 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 0.1636 \, \text{g} \). Rounding to 2 significant figures, it's \( 0.16 \, \text{g} \) (or if we consider vapor pressure of water at 50°C, vapor pressure of water at 50°C is about 0.1217 atm, so partial pressure of \( CO_2 \) is \( 1 - 0.1217 = 0.8783 \, \text{atm} \). Then recalculate \( n = \frac{0.8783 \times 0.0986}{0.0821 \times 323.15} \approx \frac{0.0865}{26.53} \approx 0.00326 \, \text{mol} \), mass \( = 0.00326 \times 44.01 \approx 0.143 \, \text{g} \), still ~0.14 - 0.16 g with 2 sig figs).

Answer:

\( \approx 0.16 \, \text{g} \) (or \( 0.14 \, \text{g} \) if accounting for water vapor, but typical assumption for simplicity gives ~0.16 g)