Question

Alnico is a strongly magnetic alloy (solid mixture) of Co, Ni, Al and Fe. A 100.0 g sample of Alnico has 11.2g Al, 22.3g Ni, and 5.0g Co in it; the rest is Fe. How many moles of aluminum are there?

Answer 1

`0.423` `mol` Al give or take w/ sig figs

Explanation:

`11.2` `g` Al x (`1` `mol`/`26.98` `g`)=`0.415` `mol` Al

`22.3` `g` Ni x(`1` `mol`/`58.69` `g`)=`0.380` `mol` Ni

`5.0` `g` Co x (`1` `mol`/`58.93` `g`)=`0.0848` `mol` Co

`100` `g -11.2` `g` Al `-22.3g` `Ni -5.0` `g` Co = `61.5` `g` Fe

`61.5` `g` Fe `xx((1 mol)/55.84)=1.10` `mol` Fe

Divide each one by the smallest number of moles: cobalt at `0.0848` `mol`.

Al = `0.415/0.0848 = 4.89` `mol`
Ni = `0.380/0.0848 = 4.48` `mol`
Co =`0.0848/0.0848 = 1.00` `mol`
Fe= `1.10/0.0848= 12.97` `mol`

Since Ni is `4.48 ~~4.50`, we multiply everything by `2` to get a whole number, `9`, for Ni.

Al = `4.89` `mol xx 2= 10` `mol` Al
Ni = `4.48` `mol xx 2= 9` `mol` Ni
Co = `1.00` `mol xx 2= 2` `mol` Co
Fe= `12.97` `mol xx 2= 26` `mol` Fe

This gives us an empirical formula of `Al_10Co_2Fe_26Ni_9`.

If we add up the Fe, Co, Ni, and Al in the empirical formula, we get `2366.7` `gmol^-1` of Alnico

`100` `g` Alnico x (`1` `mol` Alnico/`2366.7` `g` Alnico) x (`10` `mol` Al/`1` `mol` Alnico) = `0.423` `mol` Al