Làm nhanh giúp em ạ.

$\rm a.n_{Fe}=\dfrac{56}{56}=1(mol)$

trong 1 mol nguyên tử Fechứa $\rm 6,022.10^{23}$ nguyên tử

$\rm p=6,022.10^{23}.26=156,572.10^{23} ( hạt )$

$\rm n=6,022.10^{23}.30=180,66.10^{23} (hạt)$

$\rm e=6,022.10^{23}.26=156,572.10^{23}(hạt)$

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b. ta có $\rm m_{e} có trong 56g Fe =156,572.10^{23} . 9,11.10^{-31}=1,43.10^{-5} (kg)$

cứ $\rm 1,43.10^{-5}(kg) electron$ thì chứa 0,056kg Fe

thì : $\rm 1kg electron $ chứa $\rm \dfrac{0,056.1}{1,43.10^{-5}}=3916(kg)$