Question

Use the following equation: 2C6H6 + 9 O2 → 12 CO + 6 H2O (DELTA)H = -3301 kJ

If you have 1.34 g of the organic fuel (MW = 78.01 g/mol) and 3.34 g of oxygen gas, how many grams of carbon monoxide will you produce?

If you produced 345 kJ of energy, how many grams of the organic fuel was used?

Answer 1

    2C6H6 + 9 O2 ---> 12 CO + 6 H2O ; DH = -3301 kj

no of mol of C6H6 = 1.34/78.01 = 0.017 mol

no of mol of o2 = 3.34/32 = 0.104 mol

limiting reactant = C6H6

no of mol of CO produce = 0.017*12/2 = 0.102 mol

amount of CO produce = n*Mwt

                     = 0.102*28

                     = 2.856 g

no of mol of fuel required = 345*2/3301 = 0.21 mol

amount of C6H6 required = 0.21*78.01 = 16.4 g