What mass of carbon is needed to produce 25.0 L of ethene, C2H4, at 20.0 °C and 85.0 kPa according to the chemical equation given below? 2 C (s) + 2 H2 (g) → C2H4 (g)

Question

a.

20.9 g

b. 

48.8 g

c. 

154 g

d. 

307 g

Answer 1

First calculate the number of moles of C2H4 produced by the reaction, PV = nRT.

n = PV/RT = 85 kPa x 25L / (8.314 L-kPa/mole-deg x 293 °K) = 0.872 moles of C2H4.

As the balance equation shows, 2 moles of C are needed for every mole of C2H4 produced, so the moles of carbon needed is 2 x 0.872 = 1.744 moles C.

The atomic mass of C is 12.011 g/mole, so the grams of carbon needed is 12.011 g/mole x 1.744 moles = 20.9 grams to 3 sig fig.

Note that the gas constant used can be found on the internet for volume in liters and pressure in kPa.