Question

Answer 1

1.\(n_{SO_2}=\dfrac{4,48}{22,4}=0,2mol\)

\(4FeS_2+11O_2\rightarrow\left(t^o\right)2Fe_2O_3+8SO_2\)

   0,1                                               0,2       ( mol )

\(n_{FeS_2}=\dfrac{0,1}{75\%}=0,133mol\)

\(m_{FeS_2}=0,133.120=15,96g\)

\(m_{quặng}=15,96:80\%=19,95g\)

 

Answer 2

\(1,n_{SO_2} = \dfrac{4,48}{22,4} = 0,2 (mol)\\ \)

\(n_{SO_2\left(tt\right)}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{SO_2\left(lt\right)}=\dfrac{0,2}{75\%}=\dfrac{4}{15}\left(mol\right)\\ PTHH:4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\uparrow\\ Theo.pt:n_{FeS_2}=\dfrac{1}{2}n_{SO_2}=\dfrac{1}{2}.\dfrac{4}{15}=\dfrac{2}{15}\left(mol\right)\\ m_{FeS_2}=\dfrac{2}{15}.120=16\left(g\right)\\ m_{quặng}=\dfrac{16}{100\%-20\%}=20\left(g\right)\)

\(2,n_{O_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ m_{O_2}=0,03.32=0,96\left(g\right)\\ n_{O_2}=0,03.2=0,06\left(mol\right)\)

Áp dụng đkbtkl, ta có:

m_muối = m_{chất rắn} + m_O2

=> m_{chất rắn} = 2,45 - 0,96 = 1,49 (g)

\(n_K=\dfrac{1,49.52,35\%}{39}=0,02\left(mol\right)\\ n_{Cl}=\dfrac{1,49-0,02.39}{35,5}=0,02\left(mol\right)\)

CTDGN: K_xCl_yO_z

=> x : y : z = 0,02 : 0,02 : 0,06 = 1 : 1 : 3

CTDGN: KClO₃