Bài 1:

a,

$n_{Fe}=\dfrac{1000}{56}=\dfrac{125}{7}(mol)$

$\to n_e=26n_{Fe}=\dfrac{3250}{7}(mol)$

Số hạt electron: $N_e=\dfrac{3250}{7}.6,022.10^{23}=2,7959.10^{26}$

$\to m_e=9,1.10^{-28}N_e=0,254(kg)$

b,

$m_e=1000g$

$\to$ số electron $=\dfrac{1000}{9,1.10^{-31}}=1,0989.10^{33}$

$\to n_e=\dfrac{1,0989.10^{33}}{6,022.10^{23}}=\text{ 1 824 809 304} (mol)=1824809,304(kmol)$

$\to n_{Fe}=\dfrac{n_e}{26}=70185(kmol)$

$\to m_{Fe}=3930360(kg)$

Bài 2:

a,

$n_{Na}=\dfrac{69}{23}=3(mol)$

$\to n_p=3.11=33(mol)$

Số hạt proton: $33.6,022.10^{23}=1,98726.10^{25}$

$\to m_p=1,98726.10^{25}.1,67.10^{-24}=33,187242g$

b,

$m_e=1000g$

$\to$ số electron $=\dfrac{1000}{9,1.10^{-31}}=1,0989.10^{33}$

$\to n_e=\dfrac{1,0989.10^{33}}{6,022.10^{23}}=\text{ 1 824 809 304}(mol)=1824809,304(kmol)$

$\to n_{Na}=\dfrac{n_e}{11}=165891,73(kmol)$

$\to m_{Na}=3815509,8kg$