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nH2SO4 = 49/98 = 0.5 (mol)
CMH2SO4 = 0.5/0.15 = 3.3 (M)
Zn + H2SO4 => ZnSO4 + H2
...........0.5.............0.5.........0.5
VH2 = 0.5 * 22.4 = 11.2 (l)
CMZnSO4 = 0.5 / 0.15 = 10/3 (M)
C%ZnSO4 = CM*M / 10D = 10/3 * 161 / 10 * 1.25 = 42.9 %
\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\
n_{H_2SO_4}=\dfrac{24,5}{98}=0,25\left(mol\right)\\
pthh:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
\(LTL:\dfrac{0,2}{1}< \dfrac{0,25}{1}\)
=> H2SO4 dư
\(n_{H_2}=n_{H_2SO_4\left(p\text{ư}\right)}=n_{Fe}=0,2\left(mol\right)\\
V_{H_2}=0,2.22,4=4,48l\\
m_{H_2SO_4\left(d\right)}=\left(0,25-0,2\right).98=4,9g\)
a)Đổi \(V_{H_2SO_4}=100ml=0,1l\)
Số mol của 2,7 gam Al:
\(n_{Al}=\dfrac{m}{M}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)3+3H_2\)
Tỉ lệ 2 : 3 : 1 : 3
0,1 -> 0,15 : 0,05 : 0,15(mol)
Nồng độ mol của dung dịch H2SO4:
\(C_{M_{H_2SO_4}}=\dfrac{n_{H_2SO_4}}{V_{H_2SO_4}}=\dfrac{0,15}{0,1}=1,5\left(M\right)\)
b) thể tích của 0,15 mol H2:
\(V_{H_2}=n.22,4=0,15.22,4=3,36\left(l\right)\)
c) nồng độ mol của dd \(Al_2\left(SO_4\right)_3\) :
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{n}{V}=\dfrac{0,05}{0,1}=0,5\left(M\right)\)
nMg = 4,8 : 24 = 0,2 mol
a) Mg + H2SO4 → MgSO4 + H2
Theo tỉ lệ phản ứng => nH2SO4 phản ứng = nMgSO4 = nH2 = 0,2 mol
=> VH2 = 0,2.22,4 = 4,48 lít.
b)
mH2SO4 phản ứng = 0,2.98 = 19,6 gam
=> C% H2SO4 = \(\dfrac{19,6}{300}.100\text{%}\) = 6,53%
c) mMgSO4 = 0,2.120 = 24 gam.
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{FeCl_2}=n_{H_2}=n_{Fe}=0,15\left(mol\right)\\ n_{HCl}=0,15.2=0,3\left(mol\right)\\ a,m_{FeCl_2}=127.0,15=19,05\left(g\right)\\ b,V_{H_2\left(đktc\right)}=0,15.22,4=3,36\left(l\right)\\ c,C_{MddHCl}=\dfrac{0,3}{0,02}=15\left(M\right)\)
\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)
\(n_{HCl}=\dfrac{182.5\cdot10}{100\cdot36.5}=0.5\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(0.2......0.4..........0.2........0.2\)
\(n_{HCl\left(dư\right)}=0.5-0.4=0.1\left(mol\right)\)
\(m_{HCl\left(dư\right)}=0.1\cdot36.5=3.65\left(g\right)\)
\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)
\(m_{\text{dung dịch sau phản ứng}}=13+182.5-0.2\cdot2=195.1\left(g\right)\)
\(C\%_{HCl\left(dư\right)}=\dfrac{3.65}{195.1}\cdot100\%=1.87\%\)
\(C\%_{ZnCl_2}=\dfrac{0.2\cdot136}{195.1}\cdot100\%=13.94\%\)
Fe+H2SO4->FeSO4+H2
0,15---0,15-----0,15---0,15 mol
n Fe=8,4\56=0,15 mol
=>VH2=0,15.22,4=3,36l
=>m H2SO4=0,15.98=14,7g
=>C% H2SO4=14,7\245 .100=6%
=>m dd muối=8,4+245-0,15.2=253,1g
=>C% muối =0,15.152\253,1 .100=9%
\(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\\ m_{HCl}=\dfrac{100.18,25}{100}=18,25\left(g\right)\\
n_{HCl}=\dfrac{18,25}{36,5}=0,5\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,125 0,125 (mol )
\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(l\right)\\
\)
\(C\%=\dfrac{8,125}{8,125+18,25}.100\%=30,8\%\)
Bài 18:
Ta có: \(n_{Zn}=\dfrac{8,125}{65}=0,125\left(mol\right)\)
\(m_{HCl}=100.18,25\%=18,25\left(g\right)\Rightarrow n_{HCl}=\dfrac{18,25}{36,5}=0,5\left(mol\right)\)
a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, Xét tỉ lệ: \(\dfrac{0,125}{1}< \dfrac{0,5}{2}\), ta được HCl dư.
Theo PT: \(n_{H_2}=n_{Zn}=0,125\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,125.22,4=2,8\left(g\right)\)
\(m_{H_2}=0,125.2=0,25\left(g\right)\)
c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnCl_2}=n_{Zn}=0,125\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Zn}=0,25\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl\left(dư\right)}=0,25\left(mol\right)\)
Có: m dd sau pư = 8,125 + 100 - 0,25 = 107,875 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,125.136}{107,875}.100\%\approx15,76\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,25.36,5}{107,875}.100\%\approx8,46\%\end{matrix}\right.\)
Bạn tham khảo nhé!
nAl= 0,04(mol)
PTHH: 2 Al + 3 H2SO4 -> Al2(SO4)3 + 3 H2
0,04___________0,06___0,02_____0,06(mol)
a) V(H2, đktc)=0,06.22,4=1,344(l)
b) VddH2SO4= 0,06/2=0,03(l)=30(ml)
c) VddAl2(SO4)3=VddH2SO4=0,03(l)
=>CMddAl2(SO4)3=0,02/0,03=2/3(M)
\(n_{Al}=\dfrac{1.08}{27}=0.04\left(mol\right)\)
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
\(0.04......0.06.............0.02...........0.06\)
\(V_{H_2}=0.06\cdot22.4=1.344\left(l\right)\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.06}{2}=0.03\left(l\right)\)
\(C_{M_{Al_2\left(SO_4\right)_3}}=\dfrac{0.02}{0.03}=\dfrac{2}{3}\left(M\right)\)
mH2SO4=9,8g=>nH2SO4=0,1mol
nFe=0,15mol
PTHH: Fe+H2SO4=> FeSO4+H2
0,15:0,1 => n Fe dư theo N H2SO4
p/ư: 0,1<-0,1-------->0,1--->0,1
=> V H2=0,1.22,4=2,24ml
mFeSO4=0,1.152=15,2g
mdd FeSO4= 8,4+49-0,1.2=57,2g
( theo định luật bảo toàn khối lượng)
=> C%FeSO4=\(\frac{15,2}{57,2}.100=26,6\%\)