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a) 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
b) \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
PTHH: 2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
____\(\dfrac{4}{15}\)<----0,4<--------------------0,4
=> \(m_{Al}=\dfrac{4}{15}.27=7,2\left(g\right)\)
c) \(C_{M\left(H_2SO_4\right)}=\dfrac{0,4}{0,15}=2,667M\)
Theo bài ra, ta có: \(m_{Ag}=5,6\left(g\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
a) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Al}=\dfrac{1}{15}\left(mol\right)\) \(\Rightarrow m_{Al}=\dfrac{1}{15}\cdot27=1,8\left(g\right)\)
\(\Rightarrow\%m_{Al}=\dfrac{1,8}{1,8+5,6}\cdot100\%\approx24,32\%\) \(\Rightarrow\%m_{Ag}=75,68\%\)
b) Theo PTHH: \(n_{H_2SO_4}=n_{H_2}=0,1mol\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
c) PTHH: \(H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+2H_2O\)
Theo PTHH: \(n_{Ba\left(OH\right)_2}=n_{H_2SO_4}=0,1mol\)
\(\Rightarrow V_{ddBa\left(OH\right)_2}=\dfrac{0,1}{0,2}=0,5\left(l\right)=500\left(ml\right)\)
\(n_{H_2}=\dfrac{V_{H_2}}{22,4}=\dfrac{3,36}{22,4}=0,15mol\)
a. PTHH: Fe + H2SO4 \(\rightarrow\) FeSO4 + H2
TL: 1 1 1 1
mol: 0,15 \(\leftarrow\) 0,15 \(\leftarrow\) 0,15 \(\leftarrow\) 0,15
\(b.m_{Fe}=n.M=0,15.56=8,4g\)
Đổi 150ml = 0,15 l
\(c.C_{MddH_2SO_4}=\dfrac{n}{V}=\dfrac{0,15}{0,15}=1M\)
a, \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b, Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{4}{15}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{\dfrac{4}{15}.27}{10}.100\%=72\%\\\%m_{Cu}=28\%\end{matrix}\right.\)
c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{39,2}{300}.100\%\approx13,067\%\)
thôi thì mình làm cho bn vậy, câu a ko làm dc đâu, làm câu b thôi, làm sao biết dc chất nào dư khi chỉ có số mol 1 chất?
nK2SO3=0.1367(mol)
mddH2SO4=Vdd.D=200.1,04=208(g)
K2SO3+H2SO4-->K2SO4+H2O+SO2
0.1367----0.1367----0.1367---------0.1367 (mol)
mddspu=100+208-0,1367.64=299.2512(g) ; mK2SO4=0,1367.174=23.7858(g)
==>C%=23.7858.100/299.512=7.94%
2)pt bn tự ghi nhé
ta có hệ pt: 56a+27b=11 và a+3b/2=8.96/22.4==>a=0.1, b=0.2
==>%Fe=0.1x56x100/11=50.9%
%Al=100%-50.9%=49.1%
b)nH2SO4= 0.7(mol)==>VddH2SO4=0.7/2=0.35(L)
\(n_{H_2}=\dfrac{3,36}{22,4}0,15(mol)\\ a,PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ b,n_{Fe}=n_{H_2}=0,15(mol)\\ \Rightarrow \%_{Fe}=\dfrac{0,15.56}{14,8}.100\%=56,76\%\\ \Rightarrow \%_{Cu}=100\%-56,76\%=43,24\%\\ c,n_{H_2SO_4}=0,15(mol)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{0,15.98}{20\%}=73,5(g)\\ \Rightarrow V_{dd_{H_2SO_4}}=\dfrac{73,5}{1,4}=52,5(l)\)
Câu 1:
Đặt \(n_{Al}=x(mol);n_{Fe}=y(mol)\Rightarrow 27x+56y=0,83(1)\)
\(n_{H_2}=\dfrac{0,56}{22,4}=0,025(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow 1,5x+y=0,025(2)\\ (1)(2)\Rightarrow x=y=0,01(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,01.27}{0,83}.100\%=32,53\%\\ \Rightarrow \%_{Fe}=100\%-32,53\%=67,47\%\)
Câu 2:
Đặt \(n_{Al}=x(mol);n_{Mg}=y(mol)\Rightarrow 27x+24y=4,5(1)\)
\(n_{H_2}=\dfrac{5,04}{22,4}=0,225(mol)\\ 2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2\\ Mg+H_2SO_4\to MgSO_4+H_2\\ \Rightarrow 1,5x+y=0,225(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,075(mol)\\ \Rightarrow \%_{Al}=\dfrac{0,1.27}{4,5}.100\%=60\%\\ \Rightarrow \%_{Mg}=100\%-60\%=40\%\)
Dạng PP hai dòng:
\(PTHH:2A+Cl_2\to 2ACl\\ \Rightarrow n_A=n_{ACl}\\ \Rightarrow \dfrac{9,2}{M_A}=\dfrac{23,4}{M_A+35,5}\\ \Rightarrow M_A=23(g/mol)\)
Vậy A là natri
a, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)
\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)
c, \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,15\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6\%}=75\left(g\right)\)
