Số mol của 0,6x10\(^{^{23}}\)nguyên tử sắt là :

\(n_{Fe=}\frac{0,6\times10^{23}}{6\times10^{23}}\)\(=0,1\)\(mol\)

Số mol của 0,6x10²³ nguyên tử sắt là :

\(n_{Fe}=\frac{0,6\times10^{23}}{6\times10^{23}}=0,1mol\)

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\(\text{a) }4NH_3+5O_2\rightarrow4NO+6H_2O\)

\(\text{b) }S+2HNO_3\rightarrow H_2SO_4+2NO\)

a) 4NH₃ + 5O₂ -> 4NO + H₂O

b) S + 2HNO₃ -> H₂SO₄ + NO