\(a,n_{FeS_2}=\dfrac{m_{FeS_2}}{M_{FeS_2}}=\dfrac{6}{120}=0,05\left(mol\right)\\ 4FeS_2+11O_2\rightarrow\left(t^o,xt\right)2Fe_2O_3+8SO_2\uparrow\\ n_{Fe_2O_3}=\dfrac{2}{4}.n_{FeS_2}=\dfrac{2}{4}.0,05=0,025\left(mol\right)\\ \Rightarrow m_{Fe_2O_3}=160.0,025=4\left(g\right)\\ n_{SO_2}=\dfrac{8}{4}.n_{FeS_2}=\dfrac{8}{4}.0,05=0,1\left(mol\right)\\ \Rightarrow m_{SO_2}=0,1.64=6,4\left(g\right)\\ \Rightarrow m_{sp}=m_{Fe_2O_3}+m_{SO_2}=4+6,4=10,4\left(g\right)\\ b,n_{O_2}=\dfrac{11}{4}.n_{FeS_2}=\dfrac{11}{4}.0,05=0,1375\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,1375.22,4=3,08\left(l\right)\\ \Rightarrow V_{kk\left(đktc\right)}=3,08.5=15,4\left(l\right)\)

\(pthh:4FeS_2+11O_2\overset{t^o}{--->}2Fe_2O_3+8SO_2\uparrow\)

a. Ta có: \(n_{FeS_2}=\dfrac{6}{120}=0,05\left(mol\right)\)

Theo pt: \(n_{O_2}=\dfrac{11}{4}.n_{FeS_2}=\dfrac{11}{4}.0,05=0,1375\left(mol\right)\) 

\(\Rightarrow m_{sản.phẩm.thu.được}=6+0,1375.32=10,4\left(g\right)\)

b. Ta có: \(V_{O_2}=0,1375.22,4=3,08\left(lít\right)\)

Mà: \(V_{O_2}=\dfrac{1}{5}V_{kk}\)

\(\Rightarrow V_{kk}=3,08.5=15,4\left(lít\right)\)

a: \(4FeS_2+11O_2\rightarrow2Fe_2O_3+8SO_2\uparrow\)

\(n_{FeS_2}=\dfrac{6}{129}\)(mol)

\(\Leftrightarrow n_{Fe_2O_3}=\dfrac{3}{129}\left(mol\right)\)

\(m_{Fe_2O_3}=\dfrac{3}{129}\cdot178=\dfrac{178}{43}\left(g\right)\)

b: \(n_{O_2}=\dfrac{2}{129}\left(mol\right)\)

\(V=n\cdot22.4=\dfrac{2}{129}\cdot22.4=\dfrac{224}{645}\left(lít\right)\)

9,6 S phải ko bn

\(n_S=\dfrac{m}{M}=\dfrac{9,6}{32}=0,3\left(mol\right)\)

\(PTHH:S+O_2-^{t^o}>SO_2\)

tỉ lệ:         1 :   1        :    1

n(mol):   0,3--->0,3---->0,3

\(V_{O_2\left(dktc\right)}=n\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\\ V_{kk}=6,72:\dfrac{1}{5}=33,6\left(l\right)\\ V_{SO_2\left(dktc\right)}=n\cdot22,4=0,3\cdot22,4=6,72\left(l\right)\)

nFeS2 = 120/120 = 1 (mol)

PTHH: 4FeS2 + 11O2 -> (t°) 2Fe2O3 + 8SO2

Mol: 1 ---> 2,75 ---> 0,5 ---> 2

VO2 = 2,75/(100% - 10%) . 22,4 = 616/9 (l)

msp = (0,5 . 160 + 8 . 64) . 80% = 437,6 (g)

nFeS2 = 120/120 = 1 (mol)

PTHH: 4FeS2 + 11O2 -> (t°) 2Fe2O3 + 8SO2

Mol: 1 ---> 2,75 ---> 0,5 ---> 2

VO2 = 2,75/(100% - 10%) . 22,4 = 616/9 (l)

msp = (0,5 . 160 + 8 . 64) . 80% = 437,6 (g)

\(n_{C_2H_2}=\dfrac{2,6}{26}=0,1\left(mol\right)\\ 2C_2H_2+5O_2\rightarrow\left(t^o\right)4CO_2+2H_2O\\ a,n_{O_2}=\dfrac{5}{2}.n_{C_2H_2}=\dfrac{5}{2}.0,1=0,25\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\\ b,n_{CO_2}=\dfrac{4}{2}.n_{C_2H_2}=\dfrac{4}{2}.0,1=0,2\left(mol\right)\\ \Rightarrow m_{CO_2}=0.2.44=8,8\left(g\right)\\ n_{H_2O}=n_{C_2H_2}=0,1\left(mol\right)\\ \Rightarrow m_{H_2O}=0,1.18=1,8\left(g\right)\\ \Rightarrow m_{sp}=m_{CO_2}+m_{H_2O}=8,8+1,8=10,6\left(g\right)\)

4FeS₂+11O₂-to>2Fe₂O₃+8SO₂

1-------------2,75-------0,5-------2 mol

n FeS2=\(\dfrac{120}{120}=1mol\)

=>V_O2=2,75.\(\dfrac{110}{100}\).32=96,8g

H=80%

=>m _Fe2O3=0,5.160.\(\dfrac{80}{100}\)=64g

- Số mol Al là: 

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\\ n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\\ 4Al+3O_2\rightarrow\left(t^o\right)2Al_2O_3\\ Vì:\dfrac{0,5}{4}>\dfrac{0,2}{2}\Rightarrow Aldư\\ \Rightarrow n_{O_2}=\dfrac{3}{2}.n_{Al_2O_3}=\dfrac{3.0,2}{2}=0,3\left(mol\right)\\ \Rightarrow V_{O_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)

PTHH :   \(4Al+3O_2\left(t^o\right)->2Al_2O_3\)        (1)

\(n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{20,4}{27.2+16.3}=0,2\left(mol\right)\)

Từ (1) ->   \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,3\left(mol\right)\)

->  \(V_{O_2\left(đktc\right)}=n.22,4=0,3.22,4=6,72\left(l\right)\)

VO2 là gì bạn nhỉ

\(n_{Al}=\dfrac{13,5}{27}=0,5\left(mol\right)\)

\(PTHH:4Al+3O_2=2Al_2O_3\)

          \(0,5........0,5.......0,5\)

           \(2..........1,5........1\)

\(\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

Vậy...........

a) PTHH : \(2Zn+O_2-t^o->2ZnO\)

b) \(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)

Theo PTHH : \(n_{O2}=\dfrac{1}{2}n_{Zn}=0,15\left(mol\right)\)

=> \(V_{O2}=0,15.22,4=3,36\left(l\right)\)

c) Theo PTHH : \(n_{ZnO}=n_{Zn}=0,3\left(mol\right)\)

=> \(m_{ZnO}=0,3.81=24,3\left(g\right)\)

vậy ...

\(\begin{array}{l} a,\ PTHH:2Zn+O_2\xrightarrow{t^o} 2ZnO\\ b,\\ n_{Zn}=\dfrac{19,5}{65}=0,3\ (mol)\\ Theo\ pt:\ n_{O_2}=\dfrac{1}{2}n_{Zn}=0,15\ (mol)\\ \Rightarrow V_{O_2}=0,15\times 22,4=3,36\ (l)\\ c,\\ Theo\ pt:\ n_{ZnO}=n_{Zn}=0,3\ (mol)\\ \Rightarrow m_{ZnO}=0,3\times 81=24,3\ (g)\end{array}\)