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\(n_{FeS_2}=\dfrac{10^5}{120}\left(mol\right)\)
\(4FeS_2+11O_2\underrightarrow{t^0}2Fe_2O_3+8SO_2\)
\(n_{SO_2}=2n_{FeS_2}=2\cdot\dfrac{10^5}{120}=\dfrac{10^5}{60}\left(mol\right)\)
\(m_{SO_2}=\dfrac{10^5}{60}\cdot64=1.06\cdot10^5\left(g\right)=106\left(kg\right)\)
\(V_{kk}=5V_{O_2}=5\cdot\dfrac{11}{4}\cdot\dfrac{10^5}{120}\cdot22.4=256666\left(l\right)\)
\(m_{Fe_2O_3}=\dfrac{10^5}{240}\cdot160=0.6\cdot10^5\left(g\right)=90\left(kg\right)\)
a) Ta có: \(n_{FeS_2}=\dfrac{100}{120}=\dfrac{5}{6}\left(kmol\right)\)
\(\Rightarrow n_{SO_2}=\dfrac{5}{3}\left(kmol\right)\) \(\Rightarrow m_{SO_2}=\dfrac{5}{3}\cdot64\approx106,67\left(kg\right)\)
b) Theo PTHH: \(n_{O_2}=\dfrac{11}{4}n_{FeS_2}=\dfrac{55}{24}\left(kmol\right)=\dfrac{6875}{3}\left(mol\right)\)
\(\Rightarrow V_{O_2}=\dfrac{6875}{3}\cdot22,4=\dfrac{154000}{3}\left(l\right)\)
Mà Oxi chiếm khoảng 20% thể tích không khí
\(\Rightarrow V_{kk}=\dfrac{\dfrac{154000}{3}}{20\%}\approx256666,7\left(l\right)\)
c) Theo PTHH: \(n_{Fe_2O_3}=\dfrac{1}{2}n_{FeS_2}=\dfrac{5}{12}\left(kmol\right)\)
\(\Rightarrow m_{Fe_2O_3}=\dfrac{5}{12}\cdot160=66,67\left(kg\right)\)
a)
$4FeS_2 + 11O_2 \xrightarrow{t^o} 2Fe_2O_3 + 8SO_2$
b)
$n_{SO_2} = \dfrac{8,96}{22,4} = 0,4(mol)$
Theo PTHH :
$n_{FeS_2} = \dfrac{1}{2}n_{SO_2} = 0,2(mol)$
$m = 0,2.120 = 24(gam)$
a,\(n_{SO_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: FeS2 + O2 → Fe2O3 + SO2
Mol: 0,4 0,4
\(m_{FeS_2}=0,4.120=48\left(g\right)\)
\(n_S=\dfrac{3.2}{32}=0.1\left(mol\right)\)
\(n_{O_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)
\(S+O_2\underrightarrow{t^0}SO_2\)
\(0.05.0.05...0.05\)
\(\Rightarrow Sdư\)
\(V_{SO_2}=0.05\cdot22.4=1.12\left(l\right)\)
\(b.\)
\(S+O_2\underrightarrow{t^0}SO_2\)
\(0.1..0.1\)
\(V_{kk}=5V_{O_2}=5\cdot0.1\cdot22.4=11.2\left(l\right)\)
a, PT: \(S+O_2\underrightarrow{t^o}SO_2\)
Ta có: \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,05}{1}\), ta được S dư.
Theo PT: \(n_{SO_2}=n_{O_2}=0,05\left(mol\right)\) \(\Rightarrow V_{SO_2}=0,05.22,4=1,12\left(l\right)\)
b, Theo PT: \(n_{O_2}=n_S=0,1\left(mol\right)\)
\(\Rightarrow V_{O_2}=0,1.22,4=2,24\left(l\right)\)
\(\Rightarrow V_{kk}=2,24.5=11,2\left(l\right)\)
a) S + O2 --to--> SO2
b) \(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)
S + O2 --to--> SO2
0,1->0,1----->0,1
=> VSO2 = 0,1.22,4 = 2,24 (l)
c) VO2 = 0,1.22,4 = 2,24 (l)
d) \(V_{kk}=\dfrac{2,24.100}{21}=10,667\left(l\right)\)
a)PTHH:2KClO\(_3\)➞\(^{t^o}\)2KCl+3O\(_2\)
b) n\(_{KClO_3}\)=\(\dfrac{m_{KClO_3}}{M_{KClO_3}}\)=\(\dfrac{12,15}{122,5}\)\(\approx\)0,1(m)
PTHH : 2KClO\(_3\) ➞\(^{t^o}\) 2KCl + 3O\(_2\)
tỉ lệ : 2 2 3
số mol : 0,1 0,1 0,15
V\(_{O_2}\)=n\(_{O_2}\).22,4=0,15.22,4=3,36(l)
c)PTHH : 2Zn + O\(_2\) -> 2ZnO
tỉ lệ : 2 1 2
số mol :0,3 0,15 0,3
m\(_{Zn}\)=n\(_{Zn}\).M\(_{Zn}\)=0,3.65=19,5(g)
nS = 1,92/32 = 0,06 (mol)
PTHH: S + O2 -> (t°) SO2
Mol: 0,06 ---> 0,06 ---> 0,06
VSO2 (LT) = 0,06 . 22,4 = 1,344 (l)
VSO2 (TT) = 1,344 . 90% = 1,2096 (l)
\(n_{O2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Pt : \(S+O_2\underrightarrow{t^o}SO_2|\)
1 1 1
0,15 0,15 0,15
a) \(n_S=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
⇒ \(m_S=0,15.32=4,8\left(g\right)\)
b) \(n_{SO2}=\dfrac{0,15.1}{1}=0,15\left(mol\right)\)
\(V_{SO2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)
Chúc bạn học tốt
a, \(4Al+3O_2\underrightarrow{t^o}2Al_2O_3\)
b, \(n_{Al_2O_3}=\dfrac{20,4}{102}=0,2\left(mol\right)\)
Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{Al_2O_3}=0,3\left(mol\right)\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)
c, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=33,6\left(l\right)\)
\(n_{FeS_2}=\dfrac{12}{120}=0,1\left(mol\right)\)
PT: 4FeS2 + 11O2 → 2Fe2O3 + SO2
mol 0,1 → 0,275 0,05 0,025
a) \(m_{Fe_2O_3}=0,05.160=8\left(g\right)\)
b) \(V_{SO_2\left(đktc\right)}=0,025.22,4=0,56\left(l\right)\)
d) Vkhông khí (đktc) = (0,275.22,4).5 = 30,8 (g)