Consider the combustion reaction for acetylene. 2c2h2(l) + 5o2(g) mc001-1.jpg 4co2(g) + 2h2o(g) if the acetylene tank contains 37.0 mol of c2h2 and the oxygen tank contains 81.0 mol of o2, what is the limiting reactant for this reaction?

First we need to ensure that the reaction is balanced

It is balanced, so we use stochiometry to figure the amount of product formed by each individual reactant. For instance, we find by each reactant.

from

from

The oxygen will be finished when 64.8mol of have been produced. There will be excess acetylene at the end of the reaction. Hence Oxygen is the limiting reactant.

Oxygen

Explanation:

As can be seen from the balanced chemical equation,  2 moles of acetylene react with 5 moles of oxygen.

Thus 37 moles of acetylene will react with of oxygen.

But available is only 81.0 moles.

Thus now we take it the other way:

5 moles of oxygen  react with 2 moles of acetylene

81 moles of oxygen will react with of acetylene.

Limiting reagent is the reagent which limits the formation of product. Excess reagent is one which is in excess and thus remains unreacted.

Thus oxygen is the limiting reagent and acetylene is the excess reagent as (37-32.4)= 4.6 moles of acetylene are left.

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation: