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a) Gọi số mol Ca, CaCO3 là a, b (mol)
=> 40a + 100b = 19 (1)
\(m_{HCl}=\dfrac{500.4,38}{100}=21,9\left(g\right)\)
PTHH: Ca + 2HCl --> CaCl2 + H2
a--->2a------->a----->a
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
b------>2b------>b------>b
=> \(\overline{M}_Y=\dfrac{2a+44b}{a+b}=13,6.2=27,2\left(g/mol\right)\)
=> 25,2a = 16,8b (2)
(1)(2) => a = 0,1 (mol); b = 0,15 (mol)
\(\left\{{}\begin{matrix}m_{Ca}=0,1.40=4\left(g\right)\\m_{CaCO_3}=0,15.100=15\left(g\right)\end{matrix}\right.\)
b)
mdd sau pư = 19 + 500 - 0,1.2 - 0,15.44 = 512,2 (g)
mHCl(dư) = 21,9 - 36,5(2a + 2b) = 3,65 (g)
mCaCl2 = 111(a + b) = 27,75 (g)
\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{27,75}{512,2}.100\%=5,418\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{512,2}.100\%=0,713\%\end{matrix}\right.\)
a) Gọi số mol Ca, CaCO3 là a, b (mol)
=> 40a + 100b = 19 (1)
\(m_{HCl}=\dfrac{500.4,38}{100}=21,9\left(g\right)\)
PTHH: Ca + 2HCl --> CaCl2 + H2
a--->2a------->a----->a
CaCO3 + 2HCl --> CaCl2 + CO2 + H2O
b------>2b------>b------>b
=> \(\overline{M}_Y=\dfrac{2a+44b}{a+b}=13,6.2=27,2\left(g/mol\right)\)
=> 25,2a = 16,8b (2)
(1)(2) => a = 0,1 (mol); b = 0,15 (mol)
\(\left\{{}\begin{matrix}m_{Ca}=0,1.40=4\left(g\right)\\m_{CaCO_3}=0,15.100=15\left(g\right)\end{matrix}\right.\)
b)
mdd sau pư = 19 + 500 - 0,1.2 - 0,15.44 = 512,2 (g)
mHCl(dư) = 21,9 - 36,5(2a + 2b) = 3,65 (g)
mCaCl2 = 111(a + b) = 27,75 (g)
\(\left\{{}\begin{matrix}C\%_{CaCl_2}=\dfrac{27,75}{512,2}.100\%=5,418\%\\C\%_{HCl\left(dư\right)}=\dfrac{3,65}{512,2}.100\%=0,713\%\end{matrix}\right.\)
a,Fe + 2HCl → FeCl + H2 (1)
FeO + 2HCl → FeCl + H2O (2)
nH2 = 3,36/ 22,4 = 0,15 ( mol)
Theo (1) nH2 = nFe = 0,15 ( mol)
mFe = 0,15 x 56 = 8.4 (g)
m FeO = 12 - 8,4 = 3,6 (g)
a, \(n_{H_2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(Fe+2HCl->FeCl_2+H_2\left(1\right)\)
\(FeO+2HCl->FeCl_2+H_2O\left(2\right)\)
theo (1) \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
=> \(m_{Fe}=0,15.56=8,4\left(g\right)\)
=> \(m_{FeO}=12-8,4=3,6\left(g\right)\)
ta thấy : nFe =nH2 = 0,15
=> mFe =0,15 x 56 = 8,4g
%Fe=8,4/12 x 100 = 70%
=>%FeO = 100 - 70 = 30%
b) BTKLra mdd tìm mct of HCl
c) tìm mdd sau pứ -mH2 nha bạn
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Fe}=y\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo pt: \(\Rightarrow\left\{{}\begin{matrix}3x+y=0,2\\27x+56y=5,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{19}{470}\\y=\dfrac{37}{470}\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{\dfrac{19}{470}\cdot27}{5,5}\cdot100\%=19,84\%\)
\(\%m_{Fe}=100\%-19,84\%=80,16\%\)
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(MgO+2HCl\rightarrow MgCl_2+H_2O\)
a) Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)=n_{Mg}\)
\(\Rightarrow\%m_{Mg}=\dfrac{0,5\cdot24}{16}=75\%\) \(\Rightarrow\%m_{MgO}=25\%\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Mg}=0,5\left(mol\right)\\n_{MgO}=\dfrac{16\cdot25\%}{40}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{HCl}=2n_{Mg}+2n_{MgO}=1,2\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{1,2\cdot36,5}{20\%}=219\left(g\right)\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{H_2}=0,5\left(mol\right)\\n_{MgCl_2}=0,6\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{H_2}=0,5\cdot2=1\left(g\right)\\m_{MgCl_2}=0,6\cdot95=57\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{hhA}+m_{ddHCl}-m_{H_2}=234\left(g\right)\) \(\Rightarrow C\%_{MgCl_2}=\dfrac{57}{234}\cdot100\%\approx24,36\%\)
Cho mình hỏi ở cái PTHH ấy! sao ta không tính số mol ở dưới??
\(a,m_{rắn}=m_{Cu}=2,7\left(g\right)\\ \Rightarrow m_{\left(Zn,Fe\right)}=12-2,7=9,3\left(g\right)\\ n_{H_2}=0,15\left(mol\right),n_{axit}=2.0,2=0,4\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,15}{1}< \dfrac{0,4}{1}\Rightarrow axit.dư\\ \Rightarrow\left\{{}\begin{matrix}65+56b=9,3\\a+b=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{2,7}{12}.100=22,5\%\\ \%m_{Zn}=\dfrac{0,1.65}{12}.100\approx54,167\%\\ \%m_{Fe}=\dfrac{0,05.56}{12}.100\approx23,333\%\)
\(b,ddA:FeCl_2,ZnCl_2,H_2SO_4\left(dư\right)\\ m_{ddH_2SO_4}=200.1,14=228\left(g\right)\\ m_{ddA}=m_{\left(Zn,Fe\right)}+m_{ddH_2SO_4}-m_{H_2}=9,3+228-0,15.2=237\left(g\right)\)
\(C\%_{ddZnCl_2}=\dfrac{136.0,1}{237}.100\approx5,738\%\\ C\%_{ddFeCl_2}=\dfrac{127.0,05}{237}.100\approx2,679\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,15\right).98}{237}.100\approx10,338\%\)
Đã sửa lần cuối lúc 20:45
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 \(\left(1\right);\left(2\right)\Rightarrow80x+81y=12,1\left(g\right)\Rightarrow x=0,05\left(mol\right)\)
\(2x+2y=0,3\Rightarrow y=0,1\left(mol\right)\)
\(a,\) \(\%CuO=\dfrac{0,05.80.100}{12,1}=33,06\%\)
\(\%ZnO=\dfrac{0,1.80.100}{12,1}=66,94\%\)
\(b,\) \(CuO+H2SO4\rightarrow CuOSO4+H2O\)
\(0,05\rightarrow0,05\)
\(ZnO+H2SO4\rightarrow ZnSO4+H2O\)
\(0,1\rightarrow0,1\)
\(nH2SO4=0,05+0,1=0,15\left(mol\right)\)
\(mH2SO4=0,15.98=14,7\left(g\right)\)
\(mddH2SO4=14,7:20=73,5\left(g\right)\)