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 )
\(n_{FeS_2}=\dfrac{0,1}{75\%}=0,133mol\)
\(m_{FeS_2}=0,133.120=15,96g\)
\(m_{quặng}=15,96:80\%=19,95g\)
\(1,n_{SO_2} = \dfrac{4,48}{22,4} = 0,2 (mol)\\ \)
\(n_{SO_2\left(tt\right)}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{SO_2\left(lt\right)}=\dfrac{0,2}{75\%}=\dfrac{4}{15}\left(mol\right)\\ PTHH:4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\uparrow\\ Theo.pt:n_{FeS_2}=\dfrac{1}{2}n_{SO_2}=\dfrac{1}{2}.\dfrac{4}{15}=\dfrac{2}{15}\left(mol\right)\\ m_{FeS_2}=\dfrac{2}{15}.120=16\left(g\right)\\ m_{quặng}=\dfrac{16}{100\%-20\%}=20\left(g\right)\)
\(2,n_{O_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\\ m_{O_2}=0,03.32=0,96\left(g\right)\\ n_{O_2}=0,03.2=0,06\left(mol\right)\)
Áp dụng đkbtkl, ta có:
mmuối = mchất rắn + mO2
=> mchất 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: KxClyOz
=> x : y : z = 0,02 : 0,02 : 0,06 = 1 : 1 : 3
CTDGN: KClO3
