posted by Brya .
If 1.62 mold of CS2 burns with 5.65 mol of O2, how many moles of the excess reactant will still be present when the reaction is over? the chemical equation is CS2+3O2=2SO2+CO2
CS2 + 3O2 = 2SO2 + CO2
mols CS2 = 1.62
mols O2 = 5.65
How much CS2 is needed to burn 5.65 mols O2?
That's 5.65 mols O2 x (1 mol CS2/3 mols O2) = 5.65 x 1/3 = 1.88. You don't have that much; therefore, CS2 must be the limiting reagent but let's check to make sure.
How much O2 is need to use all of the CS2?
1.62 mols CS2 x (3 mols O2/1 mol CS2) = 1.62 x 3 = 4.86 and we have enough O2.
Therefore, 1.62 mols CS2 will burn using 4.86 mols O2. How much O2 is left? That's 5.65 - 4.86 = ? mols.