Question

How much of the excess reagent remains if 24.5 grams of CoO is reacted with 2.58 grams of O2? The following reaction is unbalanced. (Co = 58.93, O = 16.00 amu)

CoO + O2 → Co2O3

Answer 1

number of mole = (given mass)/(molar mass)
number of mole of CoO = 24.5/74.93
= 0.327 mole
number of mole O2 = 2.58/32
= 0.0806 mole

reaction taking place
4CoO + O2 --> 2Co2O3
according to reaction
1 mole of O2 required 4 mole CoO
0.0806 mole of O2 required (4*0.0806) mole CoO
0.0806 mole of O2 required 0.322 mole CoO

but we have 0.327 mole of CoO
so, CoO is in excess
number of mole of CoO remains = 0.327 - 0.322
= 0.005 mole

mass of CoO remains = (number of mole of CoO remains)*(molar mass of CoO)
= 0.005*74.93
= 0.375 g

Answer : 0.375 g