Tính khối lượng dung dịch hỗn hợp hai axit HCl 14,6% và H2SO4 19,6% cần dùng để hòa tan vừa hết 15,6 gam kẽm?
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Bài 9 :
\(a) n_{Fe_2O_3} = \dfrac{3,2}{160}=0,02(mol)\\ Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O\\ n_{H_2SO_4} = 3n_{Fe_2O_3} = 0,06(mol)\\ m_{dd\ H_2SO_4} = \dfrac{0,06.98}{19,6\%} = 30(gam)\\ b) \text{Chất tan : } Fe_2(SO_4)_3\\ n_{Fe_2(SO_4)_3} = n_{Fe_2O_3} = 0,02(mol)\\ m_{Fe_2(SO_4)_3} = 0,02.400 =8(gam)\)
Câu 4 :
\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
a) Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,2 0,4
b) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(m_{Mg}=0,05.24=1.2\left(g\right)\)
\(m_{MgO}=9,2-1,2=8\left(g\right)\)
c) Có : \(m_{MgO}=8\left(g\right)\)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)
\(m_{HCl}=0,05.36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)
Chúc bạn học tốt
Bạn ơi cho mik hỏi, tại sao nH2 lại là o,o5 mol v ? 1,12/22,4 là bằng 0,1 ....vậy tại sao lại ra 0,05 v ?
a) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,2<---0,6<--------------0,3
=> \(\left\{{}\begin{matrix}\%Al=\dfrac{0,2.27}{15,6}.100\%=34,615\%\\\%Al_2O_3=\dfrac{15,6-0,2.27}{15,6}.100\%=65,385\%\end{matrix}\right.\)
b) \(n_{Al_2O_3}=\dfrac{15,6-0,2.27}{102}=0,1\left(mol\right)\)
PTHH: Al2O3 + 6HCl --> 2AlCl3 + 3H2O
______0,1--->0,6
=> nHCl = 0,6+0,6 = 1,2(mol)
=> \(V_{dd}=\dfrac{1,2}{2}=0,6\left(l\right)\)
a)
$Zn + 2HCl \to ZnCl_2 + H_2$
$ZnO + 2HCl \to ZnCl_2 + H_2O$
b)
$n_{Zn} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$m_{Zn} = 0,1.65 = 6,5(gam)$
$m_{ZnO} = 14,6 - 6,5 = 8,1(gam)$
c)
$n_{ZnO} = \dfrac{8,1}{81} = 0,1(mol)$
$n_{HCl} = 2n_{Zn} + 2n_{ZnO} = 0,4(mol)$
$\Rightarrow V_{dd\ HCl} = \dfrac{0,4}{C_{M_{HCl}}}$
a, Gọi \(\left\{{}\begin{matrix}n_{Fe_2O_3}=a\left(mol\right)\\n_{ZnO}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)
\(n_{HCl}=1,5.0,2=0,3\left(mol\right)\)
PTHH:
Fe2O3 + 6HCl ---> FeCl3 + 3H2O
a-------->6a
ZnO + 2HCl ---> ZnCl2 + H2
b----->2b
=> \(\left\{{}\begin{matrix}160a+81b=8,83\\6a+2b=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,04\left(mol\right)\\b=0,03\left(mol\right)\end{matrix}\right.\left(TM\right)\)
=> \(\left\{{}\begin{matrix}m_{Fe_2O_3}=0,04.160=6,4\left(g\right)\\m_{ZnO}=0,03.81=2,43\left(g\right)\end{matrix}\right.\)
b, PTHH:
Fe2O3 + 3H2SO4 ---> Fe2(SO4)3 + 3H2O
0,04------>0,12
ZnO + H2SO4 ---> ZnSO4 + H2O
0,03->0,03
=> \(m_{H_2SO_4}=\left(0,12+0,03\right).98=14,7\left(g\right)\)
=> \(m_{ddH_2SO_4}=\dfrac{14,7.100}{30\%}=49\left(g\right)\)
gọi \(x,y\) lần lượt là số \(mol\) của\(CuO\) và \(ZnO\)
số \(mol\) \(HCl\)
\(N=Cm.V=3.0,1=0,3\left(mol\right)\)
lập \(PTHH\) :
\(CuO+2HCl\rightarrow CuCl2+H2O\)
\(x\Rightarrow2x\)
\(ZnO+2HCl\rightarrow ZnCl2+H2O\)
\(y\Rightarrow2y\)
theo \(PTP\) , ta có :
\(2x+2y=0,3\) \(\left(1\right)\)
theo đề ra :
\(mCuO+mZnO=80x+81y=12,1\left(g\right)\) \(\left(2\right)\)
từ \(\left(1\right);\left(2\right)\Rightarrow80x+81y=12,1\left(g\right)\Rightarrow x=0,05\left(mol\right)\)
\(2x+2y=0,3\Rightarrow y=0,1\left(mol\right)\)
\(a,\) \(\%CuO=\dfrac{0,05.80.100}{12,1}=33,06\%\)
\(\%ZnO=\dfrac{0,1.80.100}{12,1}=66,94\%\)
\(b,\) \(CuO+H2SO4\rightarrow CuOSO4+H2O\)
\(0,05\rightarrow0,05\)
\(ZnO+H2SO4\rightarrow ZnSO4+H2O\)
\(0,1\rightarrow0,1\)
\(nH2SO4=0,05+0,1=0,15\left(mol\right)\)
\(mH2SO4=0,15.98=14,7\left(g\right)\)
\(mddH2SO4=14,7:20=73,5\left(g\right)\)
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{0,1.27}{5,25}.100\%\approx51,43\%\\\%m_{Al_2O_3}\approx48,57\%\end{matrix}\right.\)
b, \(n_{Al_2O_3}=\dfrac{5,25-0,1.27}{102}=0,025\left(mol\right)\)
Theo PT: \(n_{HCl}=3n_{Al}+6n_{Al_2O_3}=0,45\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{0,45.36,5}{29,2\%}=56,25\left(g\right)\)
c, \(n_{H_2SO_4}=\dfrac{1}{2}n_{HCl}=0,225\left(mol\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,225.98}{19,6\%}=112,5\left(g\right)\)
\(n_{Zn}=\dfrac{15,6}{65}=0,24\left(mol\right)\)
\(Zn+2HCl\underrightarrow{ }ZnCl_2+H_2\)
mol 1: 2 : 1 : 1
mol 0,24:0,48:0,24:0,24
\(Zn+H_2SO_4\underrightarrow{ }ZnSO_4+H_2\)
mol 1 1 1 1
mol 0,24: 0,24:0,24:0,24
\(m_{ct\left(HCl\right)}=0,24.36,5=8,76\left(g\right)\)
\(m_{ct\left(H_2SO_4\right)}=0,24.98=23,52\left(g\right)\)
\(m_{dd\left(HCl\right)}=\dfrac{8,76}{14,6\%}.100\%=60\left(g\right)\)
\(m_{dd\left(H_2SO_4\right)}=\dfrac{23,52}{19,6\text{%}}.100\%=120\left(g\right)\)
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