Which of the following should occur when placing a certain amount of tetraphosphorus solid in the same container as dihydrogen gas, if #"8.0 mols"# of #"H"_2(g)# are present?

#A)# #"1 mol"# of #"P"_4(s)# reacts.
#B)# #"32 mols"# of #"P"_4(s)# are produced.
#C)# No reaction occurs with #"H"_2(g)#.
#D)# #"16 mols"# of #"PH"_3(g)# is produced.

2 Answers
Aug 21, 2017

Answer:

Solution

Explanation:

B is invalid as 8mol of #H_2# cannot produce 16 mol of #PH_3#

C is not possible as we assume the reactants are at a condition where the reaction is possible

D is also not valid as by B

A is possible if we take #P_4# to be limiting reagent and the #H_2# to be in excess. So 8 mol of #H_2# will react with 1mol of #P_4# and give 4 mol of #PH_3# and 2 mol of unreacted #H_2#.

Aug 21, 2017

From the balanced reaction:

#"P"_4(s) + 6"H"_2(g) -> 4"PH"_3(g)#,

we begin with #"8.0 mols"# of #"H"_2(g)#.

Since we have #"6 mols H"_2# for every #"1 mol P"_4(s)#, we know that

#8.0 cancel("mols H"_2(g)) xx ("1 mol P"_4(s))/(6 cancel("mols H"_2(g)))#

#= ul("1.33 mols P"_4(s))#

will react with it. Likewise, since #"P"_4(s)# is #1:4# with #"PH"_3#, we would produce #ul("5.33 mols")# of #"PH"_3(g)#.


#A)# is wrong, as we have shown it reacts with more than #"1 mol"# of #"P"_4(s)#... It is the only answer that could potentially have been right, if we started with #"6.0 mols H"_2#... but we didn't!

#B)# is wrong, because that is far too many mols produced.

#C)# Phosphorus is clearly reactive towards #"H"_2(g)#. It looks like this:

http://www.meduniv.lviv.ua/

The bond angles (#60^@#!) are strained, making #"P"_4(s)# quite reactive...

#D)# is wrong, just as #B)# is also wrong. We found only #"5.33 mols"# produced...