Question

tinh so nguyen tu oxi can dung de tac dung du voi hon hop gom 6g C va 8gS

Answer 1

\(n_C=\dfrac{m}{M}=\dfrac{6}{12}=0,5\left(mol\right)\\ n_S=\dfrac{m}{M}=\dfrac{8}{32}=0,25\left(mol\right)\)

\(pthh:C+O_2\rightarrow CO_2\left(1\right)\\ S+O_2\rightarrow SO_2\left(2\right)\)

Theo \(pthh\left(1\right):n_{O_2\left(1\right)}=n_C=0,5\left(mol\right)\)

\(pthh\left(2\right):n_{O_2\left(2\right)}=n_S=0,25\left(mol\right)\)

\(\Rightarrow\Sigma n_{O_2}=0,5+0,25=0,75\left(mol\right)\)

\(\Rightarrow\) Số nguyên tử \(O=2\cdot n\cdot N=2\cdot0,75\cdot6\cdot10^{23}=9\cdot10^{23}\left(nguyên\text{ }tử\right)\)