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PTHH: \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot9,8\%}{98}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\dfrac{0,1}{1}>\dfrac{0,2}{3}\) \(\Rightarrow\) Fe2O3 còn dư, tính theo axit
\(\Rightarrow\left\{{}\begin{matrix}n_{Fe_2\left(SO_4\right)_3}=0,1\left(mol\right)\\n_{Fe_2O_3\left(dư\right)}=\dfrac{1}{30}\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{Fe_2\left(SO_4\right)_3}=0,1\cdot400=40\left(g\right)\\m_{Fe_2O_3\left(dư\right)}=\dfrac{1}{30}\cdot160\approx5,3\left(g\right)\end{matrix}\right.\)
\(\Rightarrow C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{40}{16+200-5,3}\cdot100\%\approx18,98\%\)
\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)
a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)
1 2 1 1
0,1 0,2 0,1
b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)
⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(C_{ddHCl}=\dfrac{7,3.100}{200}=3,65\)0/0
c) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)
⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)
Chúc bạn học tốt
\(m_{CuSO_4}=\dfrac{200\cdot32}{100}=64\left(g\right)\)\(\Rightarrow n_{CuSO_4}=\dfrac{64}{160}=0,4mol\)
\(m_{BaCl_2}=\dfrac{200\cdot10,4}{100}=20,8\left(g\right)\)\(\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1mol\)
\(CuSO_4+BaCl_2\rightarrow BaSO_4\downarrow+CuCl_2\)
0,4 0,1 0,1 0,1
b)\(m_{BaSO_4}=0,1\cdot233=23,3\left(g\right)\)
c)\(m_{CuCl_2}=0,1\cdot135=13,5\left(g\right)\)
\(\Rightarrow m_{ddsau}=200+200-13,5=386,5\left(g\right)\)
\(\Rightarrow C\%=\dfrac{23,3}{386,5}\cdot100\%=6,028\%\)
Ta có: \(n_{Ba}=\dfrac{8,22}{137}=0,06\left(mol\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{m_{H_2SO_4}}{200}.100\%=1,96\%\)
=> \(m_{H_2SO_4}=3,92\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\)
PTHH: Ba + H2SO4 ---> BaSO4↓ + H2
Ta thấy: \(\dfrac{0,06}{1}>\dfrac{0,04}{1}\)
=> Ba dư
Theo PT: \(n_{BaSO_4}=n_{H_2SO_4}=0,04\left(mol\right)\)
=> \(m_{BaSO_4}=0,04.233=9,32\left(g\right)\)
Theo PT: \(n_{H_2}=n_{H_2SO_4}=0,04\left(mol\right)\)
=> \(m_{H_2}=0,04.2=0,08\left(g\right)\)
Ta có: \(m_{dd_{BaSO_4}}=8,22+200-0,08=208,14\left(g\right)\)
=> \(C_{\%_{BaSO_4}}=\dfrac{9,32}{208,14}.100\%\approx4,48\%\)
a) \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + H2SO4 → FeSO4 + H2
Mol: 0,1 0,1 0,1
PTHH: Fe2O3 + 3H2SO4 → Fe2(SO4)3 + 3H2O
Mol: 0,05 0,05
\(m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Fe_2O_3}=13,6-5,6=8\left(g\right)\Rightarrow n_{Fe_2O_3}=\dfrac{8}{160}=0,05\left(mol\right)\)
b và c ko hiểu đề
\(n_{Fe_2O_3}=\dfrac{1,6}{160}=0,01\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{49.6\%}{98}=0,03\left(mol\right)\)
PTHH:
\(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O\)
0,01 0,03 0,01
\(C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,01.400}{1,6+49}.100\%=7,91\left(\%\right)\)
c, axit phản ứng hết
\(a)ZnO+2HCl\rightarrow ZnCl_2+H_2O\\ b)n_{ZnO}=\dfrac{8,1}{81}=0,1mol\\ n_{HCl}=\dfrac{300.3,65}{100.36,5}=0,3mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,3}{2}\Rightarrow HCl.dư\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
0,1 0,2 0,1 0,1
\(m_{ZnCl_2}=0,1.136=13,6g\\ m_{HCl.dư}=\left(0,3-0,2\right).36,5=3,65g\\ m_{H_2O}=0,1.18=1,8g\\ c)C_{\%ZnCl_2}=\dfrac{13,6}{8,1+300}\cdot100=4,41\%\\ C_{\%HCl.dư}=\dfrac{3,65}{8,1+300}\cdot100=1,18\%\)
\(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)
\(m_{HCl}=3,65\%.300=10,95\left(g\right)\Rightarrow n_{HCl}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
a) PTHH : \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)
0,1 0,3 0,1
b) Xét tỉ lệ : \(\dfrac{0,1}{1}< \dfrac{0,3}{2}\Rightarrow HCldư\)
Sau phản ứng gồm có : ZnCl2 và dd HCl dư
\(m_{ZnCl2}=0,1.136=13,6\left(g\right)\)
\(m_{HCl\left(dư\right)}=\left(0,3-0,1.2\right).36,5=3,65\left(g\right)\)
c) \(m_{ddspu}=8,1+300=308,1\left(g\right)\)
\(C\%_{ddHCldư}=\dfrac{3,65}{308,1}.100\%=1,18\%\)
\(C\%_{ZnCl2}=\dfrac{13,6}{308,1}.100\%=4,41\%\)