Question

Consider the reaction for the combustion of acetone(C3H6O)(Hint :Limiting reactant problem)

C3H6O9(I) + 4O2(g) -> 3CO2(g) + 3H2O(g)

If 300.0 grams of acetone is mixed with 600.0 grams of oxygen and reacted.

A. What is the maximum number of grams of carbon dioxide that can be formed?

B. How many water molecules could be formed?

Is there some to help me on this chemistry problem?

Answer 1

C3H6O(I) + 4O2(g) -> 3CO2(g) + 3H2O(g)

1 mole of actone react with 4 moles of O2

58g of acetone react with 4*32g of O2

300g of acetone react with = 4*32*300/58   = 662g of O2

O2 is limiting reagent

C3H6O(I) + 4O2(g) -> 3CO2(g) + 3H2O(g)

4 moles of O2 react with acetone to gives 3 moles of CO2

4*32g of O2 react with acetone to gives 3*44g of CO2

600g of O2 react with acetone to gives = 3*44*600/4*32 = 618.75g of CO2

B.C3H6O(I) + 4O2(g) -> 3CO2(g) + 3H2O(g)

    4 moles of O2 react with acetone to gives 3 moles of H2O

   4*32g of O2 react with acetone to gives 3 moles of H2O

    600g of O2 react with acetone to gives = 3*600/4*32   = 14.0625 moles of H2O

no of molescules = no of moles * 6.023*10^23

                             = 14.0625*6.023*10^23 = .5*10^24 molecules