Question
How many moles of hydrogen are in a #3.2*mol# quantity of #"ammonium hydroxide"#?
Answers
#3.2xxN_Axx5=??# Explanation:
Well,
#N_A=6.022xx10^23*mol^-1# . In one mole of#"ammonium hydroxide"# ,#NH_3*H_2O# , there are clearly 5 moles of hydrogens..........And so we take the product................
#3.2*molxx6.022xx10^23*mol^-1xx5# #"hydrogen atoms"# #=# #??# Note that ammonium hydroxide is likely better represented as
#NH_3*H_2O# than#NH_4^+HO^-# ...
#9.6xx10^24# #"atoms H"# Explanation:
We're asked to calculate the number of atoms of
#"H"# in#3.2# #"mol NH"_4"OH"# .We can do this two different ways. The first is to calculate the number of molecules of
#"NH"_4"OH"# , and then atoms of#"H"# , and the second method is to first calculate the number of moles of#"H"# , and then atoms of#"H"# .Method 1:
Using Avogadro's number (
#6.022xx10^23"particles"/"mol"# ), let's calculate the number of molecules of#"NH"_4"OH"# :
#3.2cancel("mol NH"_4"OH")((6.022xx10^23"molecules NH"_4"OH")/(1 cancel("mol NH"_4"OH")))#
#= 1.927xx10^24# #"molecules NH"_4"OH"# Now, using the fact that there are
#5# atoms of#"H"# per molecule of#"NH"_4"OH"# , we have
#"atoms H" =#
#1.927xx10^24cancel("molecules NH"_4"OH")((5"atoms H")/(1 cancel("molecule NH"_4"OH")))#
#= color(red)(9.6xx10^24# #color(red)("atoms H"# rounded to
#2# significant figures, the amount given in the problem.Method 2:
For the second method, we can calculate the number of moles of
#"H"# per mole of#"NH"_4"OH"# , and then atoms#"H"# from moles:
#3.2# #cancel("mol NH"_4"OH")((5cancel("mol H"))/(1cancel("mol NH"_4"OH")))((6.022xx10^23"atoms H")/(1cancel("mol H")))#
#= color(red)(9.6xx10^24# #color(red)("atoms H"# rounded to
#2# significant figures, the amount given in the problem.As you can see, both methods yield the same result, so either one can be used (I would probably prefer the second method; I find it to be a little more exact).