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n Zn= 19,5/65=0,3 (mol).
PTPƯ: Zn(0.3) + HCl(0.6) ----> ZnCl2(0.3) + H2(0,3)
mHCl=0,6.36.5=21.9(g)
a) C%HCl= 21.9/300.100%=7,3%
b) VH2=0,3.22,4=6,72(lít)
c) mH2=0,3.2=0,6(g)
mZnCl2=0,3.136=40,8(g)
mddZnCl2 =(19,5+300)-0,6=318,9(g)
C%=mZnCl2/mddZnCl2.100= 40,8/318,9.100=12,793%
\(a,n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,05<-0,1<------0,05<---0,05
\(b,m_{Fe}=0,05.56=2,8\left(g\right)\\ c,m_{ddHCl}=\dfrac{0,1.36,5}{14,6\%}=25\left(g\right)\\ m_{dd}=25+2,8-0,05.2=27,7\left(g\right)\\ \rightarrow C\%_{FeCl_2}=\dfrac{0,05.127}{27,7}.100\%=22,92\%\)
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\n_{H_2SO_4}=\dfrac{200\cdot4,9\%}{98}=0,1\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Cả 2 chất p/ứ hết
b+c) Theo PTHH: \(n_{ZnSO_4}=n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnSO_4}=0,1\cdot161=16,1\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Zn}+m_{ddH_2SO_4}-m_{H_2}=206,3\left(g\right)\)
\(\Rightarrow C\%_{ZnSO_4}=\dfrac{16,1}{206,3}\cdot100\%\approx7,8\%\)
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ b,n_{HCl}=2n_{H_2}=0,3(mol);n_{FeCl_2}=n_{H_2}=0,15(mol)\\ \Rightarrow m_{HCl}=0,3.36,5=10,95(g)\\ m_{FeCl_2}=0,15.127=19,05(g)\)
Bài 1:
nFe = \(\dfrac{1,12}{56}=0,02\) mol
Pt: 2Fe + 6H2SO4 (đ,n) --> Fe2(SO4)3 + 3SO2 + 6H2O
0,02 mol---------------------> 0,01 mol--> 0,03 mol
mFe2(SO4)3 = 0,01 . 400 = 4 (g)
VSO2 = 0,03 . 22,4 = 0,672 (lít)
Bài 2:
nNO = \(\dfrac{4,48}{22,4}=0,2\) mol
Pt: 3Zn + 8HNO3 (loãng) --> 3Zn(NO3)2 + 2NO + 4H2O
....0,3 mol<--------------------------------------0,2 mol
mZn pứ = 0,3 . 65 = 19,5 (g)
a)\(Fe+H2SO4-->FeSO4+H2\)
\(n_{Fe}=\frac{28}{56}=0,5\left(mol\right)\)
\(n_{H2SO4}=n_{Fe}=0,5\left(mol\right)\)
\(m_{H2SO4}=0,5.98=49\left(g\right)\)
b)\(n_{H2}=n_{Fe}=0,5\left(mol\right)\)
\(V_{H2}=0,5.22,4=11,2\left(l\right)\)
c)\(n_{FeSO4}=n_{Fe}=0,5\left(mol\right)\)
\(m_{FeSO4}=0,5.152=76\left(g\right)\)
\(a.n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Zn+FeSO_4\rightarrow ZnSO_4+Fe\\ n_{Zn}=n_{FeSO_4}=n_{Fe}=0,2mol\\ m_{Zn}=0,2.65=13g\\ b.m_{FeSO_4}=0,2.152=30,4g\\ C_{\%FeSO_4}=\dfrac{30,4}{300}\cdot100=10,13\%\)