Question

3) For the following reaction, 9.59 grams of benzene (C6H6) are allowed to react with 15.3 grams of oxygen gas . benzene (C6H6)(l) + oxygen(g) carbon dioxide(g) + water(g) What is the maximum amount of carbon dioxide that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete? grams

Answer 1

balanced equation is

2 C6H6 (l) + 15 O2 (g) .............> 12 CO2 (g) + 6 H2O (g)

9.59 grams of benzene (C6H6) = mass / molar mass = 9.59 g / 78 g / mole = 0.123 mole.

15.3 grams of oxygen gas = 15.3 / 32 =0.478 mole.

benzene remain in excess.

oxygen gas is the limiting reactant and formula is O2.

amount of carbon dioxide that can be formed = 12 * 0.478 / 15 = 0.3824 mole = 0.3824 * 44 = 16.8 gm

amount of the excess reagent = 0.123 - (2*0.478/15) = 0.059 mole = 0.059 * 78 = 4.6 gms