Question

Answer 1

a) 

3Fe + 2O₂ --t^o--> Fe₃O₄

S + O₂ --t^o--> SO₂

b)

Gọi số mol Fe, S là a, b (mol)

=> 56a + 32b = 100 (1)

PTHH: 3Fe + 2O₂ --t^o--> Fe₃O₄

              a-->\(\dfrac{2}{3}a\)

            S + O₂ --t^o--> SO₂

              b-->b

=> \(\dfrac{2}{3}a+b=\dfrac{33,6}{22,4}=1,5\) (2)

(1)(2) => a = 1,5 (mol); b = 0,5 (mol)

\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{1,5.56}{100}.100\%=84\%\\\%m_S=\dfrac{0,5.32}{100}.100\%=16\%\end{matrix}\right.\)

Answer 2

Gọi: \(\left\{{}\begin{matrix}n_S=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\\ \Rightarrow32a+56b=100\left(g\right)\left(1\right)\\ PTHH:S+O_2\underrightarrow{t^o}SO_2\\ Mol:a\rightarrow a\rightarrow a\\ 3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\\ Mol:b\rightarrow\dfrac{2b}{3}\rightarrow\dfrac{b}{3}\\ n_{O_2}=\dfrac{33,6}{22,4}=1,5\left(mol\right)\\ Hệ.pt:\left\{{}\begin{matrix}32a+56b+100\\a+\dfrac{2b}{3}=1,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\left(mol\right)\\b=1,5\left(mol\right)\end{matrix}\right.\\ m_S=0,5.32=16\left(g\right)\\ \%m_S=\dfrac{16}{100}=16\%\\ \%m_{Fe}=100\%-16\%=84\%\)