Question

Ethyne gas combusts with oxygen gas according to the following reaction: Calculate the volume, in mL of CO2 produced when 73 g of C2H2 react at 37.4 °C and 1.6 atm. (R = 0.08206 L atm/mol K) LaTeX: 2\:C_2H_2\left(g\right)\:+\:5O_2\left(g\right)\:\longrightarrow\:4\:CO_2\left(g\right)\:+2\:H_2O\left(l\right)\:

Answer 1

Answer: Volume of CO2 is 89127 mL

Explanation: The reaction that takes place is: C2H2 + O2 --> CO2 + H2O

The amount of C2H2 that react allow us to predict the amount of CO2 that will be obtained

$mol CO2 = 73gC2H2 .\frac{1 mol C2H2}{26gC2H2} . \frac{2mol CO2}{4mol C2H2} = 5,6 mol CO2$

26g/1mol is molar mass of C2H2 and 2/4 is the molar relation between CO2 and C2H2 in this reaction. Canceling units, at the end mol of CO2 are obtained

Now with the moles of CO2 and the ideal gases equation is possible to calculate the volumen occupied by the gas.

PV = RnT where P: pressure, V: volume, R: ideal gas constant, n: moles and T: temperature expressed in K (add 273,15 to °C temperature: 37,4°C + 273,15 = 310,55K)

V= RnT/P

$V= \frac{0,08206 atmL/molK . 5,6 mol. 310,55 K}{1,6 atm} =89,127 L$

To express volume in mL multiply the L result by 1000 which equals 89127 mL