Calculate the mass of the product formed by the reaction of 1940 kg of benzene with 500 m³ of ethene in the presence of anhydrous

Calculate the mass of the product formed by the reaction of 1940 kg of benzene with 500 m³ of ethene in the presence of anhydrous aluminum chloride, if 89.6% of ethene has entered the reaction.

C6H6 + C2H4 – (AlCl3) → C6H5C2H5
500 m³ = 500,000 l
⇒ n (C2H4) = (500,000 x 0.896) / 22.4 = 20,000 mol – deficient
1,940 kg = 1,940,000 g
n (C6H6) = 1 940 000 / (6 x 12 + 6) = 1 940 000/78 = 24 871.8 – in excess
⇒ n (ethylbenzene) = 20,000 x (8 x 12 + 10) = 2120 kg
Answer: the mass of the product (ethylbenzene) is 2120 kg.