a, PT: \(4FeS+7O_2\underrightarrow{t^o}2Fe_2O_3+4SO_2\)

\(4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\)

Giả sử: \(\left\{{}\begin{matrix}n_{FeS}=x\left(mol\right)\\n_{FeS_2}=y\left(mol\right)\end{matrix}\right.\)

⇒ 88x + 120y = 17,8 (1)

Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Theo PT: \(n_{SO_2}=n_{FeS}+2n_{FeS_2}=x+2y\left(mol\right)\)

⇒ x + 2y = 0,25 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,075\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{FeS}=0,1.88=8,8\left(g\right)\\m_{FeS_2}=0,075.120=9\left(g\right)\end{matrix}\right.\)

b, Theo PT: \(n_{O_2}=\dfrac{7}{4}n_{FeS}+\dfrac{11}{4}n_{O_2}=0,38125\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,38125.22,4=8,54\left(l\right)\)

Bạn tham khảo nhé!

a) Gọi n_FeS = a (mol)

       \(n_{FeS_2}=b\left(mol\right)\) với a; b > 0

\(n_{SO_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

Ta có: \(\left\{{}\begin{matrix}m_{hh}=17,8=m_{FeS}+m_{FeS_2}=88a+120b\\n_{S\left(SO_2\right)}=0,25=n_{FeS}+2n_{FeS_2}\left(bt\left[S\right]\right)=a+2b\end{matrix}\right.\)

=> a = 0,1(mol); b = 0,075(mol)

m_FeS= n.M= 0,1 . 88 = 8,8(g)

=> \(m_{FeS_2}=m_{hh}-m_{FeS}=17,8-8,8=9\left(g\right)\)

b) PT:

\(4FeS+7O_2\underrightarrow{t^o}2Fe_2O_3+4SO_2\uparrow\\ 4FeS_2+11O_2\underrightarrow{t^o}2Fe_2O_3+8SO_2\uparrow\)

\(Theo2pt\Rightarrow n_{O_2}=\dfrac{7n_{FeS}+11n_{FeS_2}}{4}=0,38125\left(mol\right)\)

\(\Rightarrow V_{O_2}=n\cdot22,4=0,38125\cdot22,4=8,54\left(l\right)\)

Câu c chất rắn là gì vậy bạn?

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(n_{O_2}=\dfrac{8.96}{22.4}=0.4\left(mol\right)\)

\(2Zn+O_2\underrightarrow{^{^{t^0}}}2ZnO\)

LTL : \(\dfrac{0.2}{2}< \dfrac{0.4}{1}\Rightarrow O_2dư\)

\(m_{O_2\left(dư\right)}=\left(0.4-0.1\right)\cdot32=9.6\left(g\right)\)

\(m_{ZnO}=0.2\cdot81=16.2\left(g\right)\)

a) \(n_{CH_4}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

            0,4---------------->0,4

=> \(V_{CO_2}=0,4.22,4=8,96\left(l\right)\)

b) \(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Xét tỉ lệ \(\dfrac{0,4}{1}>\dfrac{0,4}{2}\) => CH₄ dư, O₂ hết

PTHH: CH₄ + 2O₂ --t^o--> CO₂ + 2H₂O

                      0,4-------->0,2

=> \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

\(n_{SO_2}=\dfrac{2,8}{22,4}=0,125mol\)

\(S+O_2\rightarrow\left(t^o\right)SO_2\)

      0,125      0,125    ( mol )

\(2KMnO_4\rightarrow\left(t^o\right)K_2MnO_4+MnO_2+O_2\)

     0,25                                                  0,125    ( mol )

\(m_{KMnO_4}=0,25.158=39,5g\)

a)\(n_{Fe}=\dfrac{16,8}{56}=0,3mol\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,3                    0,3        0,3

\(V_{H_2}=0,3\cdot22,4=6,67l\)

\(m_{FeCl_3}=0,3\cdot127=38,1g\)

b)\(n_{Fe_2O_3}=\dfrac{18}{160}=0,1125mol\)

\(Fe_2O_3+3H_2\rightarrow2Fe+3H_2O\)

0,1125    0,3        0          0

0,1          0,3        0,2      0,3

0,0125     0          0,2      0,3

\(m_{Fe}=0,2\cdot56=11,2g\)

46,4 g oxit nào

46,4 gam gì vậy bạn ( câu b ý )