When 0.100 mol of carbon is burned in a close vessel with 8.00g of oxygen, how many grams of carbon dioxide can form? Which reactant is in excess, and how many grams of it remain after the reaction?

Answer 1

C + O2 ==> CO2

You have 0.1 mol C.
Convert 8 g O2 to moles. 8/32 = 0.25 mol.

0.1 mole C can form 0.1 mole CO2
0.25 mole O2 can form 0.25 mole CO2.
In limiting reagent problems, the smaller number is ALWAYS the correct one to choose. So C is the limiting reagent and will react completely. Some oxygen will remain unreacted.

g CO2 formed = 0.1 mole CO2 x molar mass CO2 = 0.1 x 44 = ?? grams CO2.

How much oxygen remains. First, determine how much oxygen is used.
0.1 mols CO2 will use 0.1 mole O2 and 0.1 mole O2 is 0.1 x 32 = 3.2 grams.
We had 8 g initially; therefore, the amount remaining is 8.0 - 3.2 = ?? grams. Check my work carefully. It's getting late here.