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nS=0.5(mol)
S+O2->SO2
Theo pthh nS=nO2=nSO2->nS=nO2=nSO2=0.5(mol)
->V1=V2=11.2(l)
a)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
0,2--->0,4---->0,2--->0,2
\(V_2=0,2.22,4=4,48\left(l\right)\)
\(V_1=\dfrac{0,4}{0,5}=0,8\left(l\right)\)
b)
\(C_{M\left(ZnCl_2\right)}=\dfrac{0,2}{0,8}=0,25M\)
c)
\(n_{H_2}=0,1\left(mol\right)\); \(n_{CuO}=\dfrac{32}{80}=0,4\left(mol\right)\)
PTHH: CuO + H2 --to--> Cu + H2O
Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,1}{1}\) => CuO dư, H2 hết
PTHH: CuO + H2 --to--> Cu + H2O
0,1<--0,1------>0,1
=> m = 32 - 0,1.80 + 0,1.64 = 30,4 (g)
PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(FeSO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+Fe\left(OH\right)_2\downarrow\)
Ta có: \(n_{H_2SO_4}=0,3\cdot0,5=0,15\left(mol\right)=n_{Fe}=n_{H_2}=n_{Ba\left(OH\right)_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Fe}=0,15\cdot56=8,4\left(g\right)\\V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\\V_{Ba\left(OH\right)_2}=\dfrac{0,15}{1}=0,15\left(l\right)=150\left(ml\right)\end{matrix}\right.\)
*Bạn xem lại đề vì nếu FeSO4 p/ứ hết thì sẽ có nhiều hơn 41,7 gam kết tủa
nO2=0,3mol
pthh: S+O2=>SO2
0,3<-0,3->0,3
=> m=0,3.32=9,6g
V=0,3.22,4=6,72l
\(2KMnO_4\underrightarrow{to}K_2MnO_4+MnO_2+O_2\\ 3Fe+2O_2\underrightarrow{to}Fe_3O_4\\ n_{Fe_3O_4}=\dfrac{69,6}{232}=0,3\left(mol\right)\\ \Rightarrow n_{O_2}=2.0,3=0,6\left(mol\right)\\ n_{KMnO_4}=2.n_{O_2}=2.0,6=1,2\left(mol\right)\\ m=m_{KMnO_4}=158.1,2=189,6\left(g\right)\\ V=V_{O_2\left(đktc\right)}=0,6.22,4=13,44\left(l\right)\)
\(a,2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ n_{KMnO_4}=\dfrac{395}{158}=2,5(mol)\\ \Rightarrow n_{O_2}=1,25(mol)\\ \Rightarrow V_{O_2}=1,25.22,4=28(l)\\ \Rightarrow V_{O_2(tt)}=28.85\%=23,8(l)\)
\(b,n_{O_2}=\dfrac{67,2}{22,4}=3(mol)\\ 2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\\ \Rightarrow n_{KMnO_4}=6(mol)\\ \Rightarrow m_{KMnO_4}=6.158=948(g)\\ \Rightarrow m_{KMnO_4(tt)}=\dfrac{948}{80\%}=1185(g)\)
PTHH: \(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\)
\(S+O_2\xrightarrow[]{t^o}SO_2\)
Ta có: \(n_{KClO_3}=\dfrac{18,375}{122,5}=0,15\left(mol\right)\) \(\Rightarrow n_{O_2\left(lý.thuyết\right)}=0,225\left(mol\right)\)
\(\Rightarrow n_{O_2\left(thực\right)}=0,225\cdot85\%=0,19125\left(mol\right)=n_S=n_{SO_2}\)
\(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,19125\cdot22,4=4,284\left(l\right)=V_{SO_2}\\m_S=0,19125\cdot32=6,12\left(g\right)\\\end{matrix}\right.\)