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a, Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH:
2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2
a---->1,5a--------------------------->1,5a
Mg + H2SO4 ---> MgSO4 + H2
b------>b----------------------->b
Hệ pt \(\left\{{}\begin{matrix}27a+24b=6,3\\1,5a+b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,15\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,1.27=2,7\left(g\right)\\m_{Mg}=0,15.24=3,6\left(g\right)\end{matrix}\right.\\ \rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{2,7}{6,3}=42,86\%\\\%m_{Mg}=100\%-42,86\%=57,14\%\end{matrix}\right.\)
b, \(n_{H_2SO_4}=0,1.1,5+0,15=0,3\left(mol\right)\)
\(\rightarrow V_{ddH_2SO_4}=\dfrac{0,3}{0,5}=0,6\left(l\right)=600\left(ml\right)\)
c, đề yêu cầu jv?
Quy hh về FeO : a mol và Fe2O3: b mol (trong từng phần)
Phần 1: mFe2O3=8,8g --> nFe2O3=0,055 mol -->a/2 +b=0,055
Phần 2: nKMnO4=0,01 mol -->n Fe2+=0,05=nFeO=a
-->b=0,03 mol
m=16,8 g ; nH2SO4=nO=0,28 mol -->V=0,56l
\(n_O=\dfrac{33,3-21,3}{16}=0,75\left(mol\right)\)
=> nH2O = 0,75 (mol)
Giả sử có V lít dd
=> \(\left\{{}\begin{matrix}n_{H_2SO_4}=V\left(mol\right)\\n_{HCl}=2V\left(mol\right)\end{matrix}\right.\)
Bảo toàn H: 2V + 2V = 0,75.2
=> V = 0,375 (lít) = 375 (ml)
PTHH: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(2Cu+O_2\underrightarrow{t^o}2CuO\)
Ta có: \(n_{O_2}=\dfrac{1,6}{32}=0,05\left(mol\right)\)\(\Rightarrow n_{Cu}=n_{CuO}=0,1\left(mol\right)\)
\(\Rightarrow\%m_{CuO}=\dfrac{0,1\cdot80}{40}\cdot100\%=20\%\)
\(\Rightarrow\%m_{Fe_2O_3}=80\%\)
CuO + 2HCl → CuCl2 + H2O (1)
Cu + 2H2SO4(đặc,nóng) → CuSO4 + SO2↑ + 2H2O (2)
CuCl2 + 2NaOH → 2NaCl + Cu(OH)2↓ (3)
\(n_{Cu\left(OH\right)_2}=\frac{39,2}{98}=0,4\left(mol\right)\)
\(n_{SO_2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)
a) Theo PT3: \(n_{CuCl_2}=n_{Cu\left(OH\right)_3}=0,4\left(mol\right)\)
Theo Pt1: \(n_{CuO}=n_{CuCl_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{CuO}=0,4\times72=28,8\left(g\right)\)
Theo pT2: \(n_{Cu}=n_{SO_2}=0,5\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,5\times64=32\left(g\right)\)
\(\%Cu=\frac{32}{32+28,8}\times100\%=52,63\%\)
\(\%CuO=100\%-52,63\%=47,37\%\)
b) Theo PT3: \(n_{NaOH}=2n_{Cu\left(OH\right)_2}=2\times0,4=0,8\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,8\times40=32\left(g\right)\)
\(\Rightarrow m_{ddNaOH}=\frac{32}{25\%}=128\left(g\right)\)
\(\Rightarrow V=V_{ddNaOH}=\frac{128}{1,28}=100\left(l\right)\)