Question

If 8.66 X 1021 atoms are removed from 0.055 mole of iron, what mass of iron remains?

Answer 1

Solution.

$\Delta N = 8.66 \sdot 10^{21} atoms;$

$\nu_0 = 0.055 mol;$

We make a proportion:

$1 mol ---- 6.02\sdot10^{23} atoms;$

$0.055 mol ---x atoms;$

$x =\dfrac{0.055mol\sdot6.02\sdot10^{23}}{1 mol} =33.11\sdot10^{21} atoms;$

$N = 33.11\sdot10^{21} - 8.66\sdot10^{21} = 24.45\sdot10^{21} atoms;$

Let's find how many moles of iron are left:

$1 mol ---- 6.02\sdot10^{23} atoms;$

$\nu mol----24.45\sdot10^{21} atoms;$

$\nu = \dfrac{24.45\sdot10^{21}atoms \sdot1mol}{6.02\sdot10^{23} atoms} = 0.041mol;$

$\nu =\dfrac{m}{M}; \implies m=\nu\sdot M;$

$m = 0.041mol\sdot56g/mol =2.27g;$

Answer: $m =2.27g;$