Question #e7ecb

The trick here is to realize that when you're working with gases kept under the same conditions for pressure and temperature, which seems to be the case here, the mole ratio that exists between the gases is equivalent to a volume ratio.

When pressure and temperature are kept constant, let's say #P# and #T#, you can use the ideal gas law equation to write

#P * V_(HF) = n_(HF) * RT -># for hydrogen fluoride

#P * V_(N_2F_2) = n_(N_2F_2) * RT -># for dinitrogen difluoride

#P * V_"product" = n_"product" * RT -># for the product of the reaction

You can thus get the mole ratio that exists between the gases by dividing these equations. For example, the two reactants will have

#(color(red)(cancel(color(black)(P))) * V_(HF))/(color(red)(cancel(color(black)(P))) * V_(N_2F_2)) = (n_(HF) * color(red)(cancel(color(black)(RT))))/(n_(N_2F_2) * color(red)(cancel(color(black)(RT))))#

which is equivalent to saying that

#color(purple)(|bar(ul(color(white)(a/a)color(black)(n_(HF)/n_(N_2F_2) = V_(HF)/V_(N_2F_2))color(white)(a/a)|))) -># the mole ratio is equivalent to the volume ratio

This means that the two reactants will react in a #2:1# mole ratio, since

#n_(HF)/n_(N_2F_2) = (10 color(red)(cancel(color(black)("cm"^3))))/(5color(red)(cancel(color(black)("cm"^3)))) = 2/1#

Notice that the reaction produces #"10 cm"^3# of gaseous product, which means that you have a #1:1# mole ratio the product and hydrogen fluoride.

You can thus say that you have

#color(red)(2)"HF"_ ((g)) + "N"_ 2"F"_(2(g)) -> color(red)(2) xx color(blue)("product")#

Now all you have to do is count the atoms that are present on the reactants' side and make sure that you add them to the products' side as well.

The reactants' side contains #2# atoms of hydrogen, #2# atoms of nitrogen, and #4# atoms of fluorine, which means that the product will contain #1# atom of hydrogen, #1# atom of nitrogen, and #2# atoms of fluorine.

The chemical equation will thus be

#color(red)(2)"HF"_ ((g)) + "N"_ 2"F"_ (2(g)) -> color(red)(2)"NHF"_(2(g))#

The product of the reaction is called difluoramine.