Asked 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
Answers
Answered by
DrBob222
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.
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.