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 55.0°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 90.5 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.

Explanation:

Step1: Find the pressure of \(CO_2\)

Assume atmospheric pressure \(P_{total}= 1\ atm\). The vapor - pressure of water at \(T = 55.0^{\circ}C\) is \(P_{H_2O}= 118.0\ mmHg\). Convert \(P_{total}\) to \(mmHg\): \(P_{total}=760\ mmHg\).
Using Dalton's law \(P_{CO_2}=P_{total}-P_{H_2O}\), so \(P_{CO_2}=760 - 118.0=642\ mmHg\). Convert to \(atm\): \(P_{CO_2}=\frac{642}{760}\ atm\approx0.845\ atm\).

Step2: Convert volume and temperature

Volume \(V = 90.5\ mL=0.0905\ L\). Temperature \(T=(55.0 + 273.15)K = 328.15\ K\).

Step3: Use the ideal gas law \(PV=nRT\)

\(R = 0.0821\ L\cdot atm\cdot K^{-1}\cdot mol^{-1}\). From \(n=\frac{PV}{RT}\), substitute \(P = 0.845\ atm\), \(V = 0.0905\ L\), \(T = 328.15\ K\), and \(R = 0.0821\ L\cdot atm\cdot K^{-1}\cdot mol^{-1}\)
\(n=\frac{0.845\times0.0905}{0.0821\times328.15}\ mol\)
\(n=\frac{0.0765}{26.94}\ mol\approx0.00284\ mol\)

Step4: Calculate the mass of \(CO_2\)

The molar mass of \(CO_2\) is \(M = 44.01\ g/mol\). Using \(m=n\times M\), \(m = 0.00284\ mol\times44.01\ g/mol\approx0.125\ g\approx0.13\ g\)

Answer:

\(0.13\ g\)