Question

How many moles of hydrogen are in a `3.2*mol` quantity of `"ammonium hydroxide"`?

Answer 1

`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^-`...

Answer 2

`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).