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PTHH ( I ) : \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\) .
PTHH ( II ) : \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\) .
\(n_{H_2}=\frac{V_{H_2}}{22,4}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
- Theo PTHH ( I ) : \(n_{Zn}=n_{ZnCl_2}=n_{H_2}=0,2\left(mol\right)\)
=> \(m_{Zn}=n.M=0,2.65=13\left(g\right)\)
=> \(m_{ZnCl_2}=n.M=0,2.\left(65+35,5.2\right)=27,2\left(g\right)\)
Mà sau phản ứng chỉ thu được \(ZnCl_2\) .
=> mmuối khan \(=m_{ZnCl_2\left(I\right)}+m_{ZnCl_2\left(II\right)}=27,2+m_{ZnCl_2\left(II\right)}=40,8\)
=> \(m_{ZnCl_2\left(II\right)}=13,6\left(g\right)\)
=> \(n_{ZnCl2}=\frac{m}{M}=\frac{13,6}{65+35,5.2}=0,1\left(mol\right)\)
Theo PTHH ( II ) : \(n_{ZnO}=n_{ZnCl_2}=0,1\left(mol\right)\)
=> \(m_{ZnO}=n.M=0,1.\left(16+65\right)=8,1\left(g\right)\)
1. Đặt kim loại hóa trị 2 là X
ta có phương trình hoá học
X+2HCl → XCl2+H2 (áp dụng quy tắc nhân chéo )
9,75_______20,4
X_______X+70
\(\frac{9,75}{X}=\frac{20,4}{X+70}\)
⇒X≈64 ( Cu)
Cu+2HCl →CuCl2 +H2
0,15 _____________ 0,15
\(n_{Cu}=\frac{9,75}{64}=0,15\left(mol\right)\)
\(\rightarrow V_y=0,15.22,4=3,36\left(l\right)\)
`1)`
`n_{Al}={2,7}/{27}=0,1(mol)`
`2Al+3H_2SO_4->Al_2(SO_4)_3+3H_2`
`0,1->0,15->0,05->0,15(mol)`
`V_{dd\ H_2SO_4}={0,15}/1=0,15(l)=150(ml)`
`->V=150`
`V'=V_{H_2}=0,15.22,4=3,36(l)`
`C_{M\ X}=C_{M\ Al_2(SO_4)_3}={0,05}/{0,15}=1/3M`
`2)`
`n_{Fe}={2,8}/{56}=0,05(mol)`
`Fe+2HCl->FeCl_2+H_2`
`0,05->0,1->0,05->0,05(mol)`
`V_{dd\ HCl}={0,1}/1=0,1(l)=100(ml)`
`->V=100`
`V_{H_2}=0,05.22,4=1,12(l)`
`C_{M\ FeCl_2}={0,05}/{0,1}=0,5M`
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
x______ 2x ____ x ______x
\(Fe+2HCl\rightarrow FeCl_2+H_2\)
y_____2y_______y______ y
\(\rightarrow\left\{{}\begin{matrix}24x+56=8\\x+y=\frac{4,48}{22,4}=0,2\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Mg}=\frac{0,1.24}{8}.100\%=30\%;\%m_{Fe}=100\%-30\%=70\%\)
\(n_{HC}=2.0,2+2.0,1=0,4\left(mol\right)\)
\(m_{Dd_{HCl}}=\frac{0,4.36,5}{14,6\%}=100\left(g\right)\)
\(C\%_{MgCl2}=\frac{0,1\left(24+71\right)}{8+100-0,2.2}=8,83\%\)
\(C\%_{FeCl2}=\frac{0,\left(56+71\right)}{8+100-0,2.2}.100\%=11,8\%\)
Đáp án A
Chất rắn Y là Cu không phản ứng
nHCl = 
= 2.0,35 = 0,7
mmuối = mKL + mgốc axit = (9,14 – 2,54) + 0,7.35,5 = 31,45(g)
\(2R+6HCl\rightarrow2RCl_3+3H_2\)
0,1___0,3____0,1______0,15
\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
m tăng=mR - mH2
\(\Leftrightarrow2,4=m_R-0,15.2\)
\(\rightarrow m_R=2,7\)
\(M_R=\frac{2,7}{0,1}=27\left(Al\right)\)
\(C\%_{HCl}=\frac{0,3.36,5}{109,5}.100\%=10\%\)
\(C\%AlCl_3=\frac{0,1.\left(27+35,5.3\right)}{2,7+109,5-0,15.2}.100\%=11,93\%\)
\(Đặt:n_{Mg}=a\left(mol\right);n_{Al}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ \Rightarrow\left\{{}\begin{matrix}24a+27b=5,1\\22,4a+22,4.1,5.b=5,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ a,\Rightarrow\%m_{Mg}=\dfrac{0,1.24}{5,1}.100\approx47,059\%\\ \Rightarrow\%m_{Al}\approx100\%-47,059\%\approx52,941\%\\ b,n_{HCl}=2.n_{H_2}=2.\left(0,1+0,1.1,5\right)=0,5\left(mol\right)\\ \Rightarrow V_{ddHCl}=\dfrac{0,5}{2}=0,25\left(l\right)\)
a)\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
x 2x x x
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
y 3y y 1,5y
Ta có hệ:
\(\left\{{}\begin{matrix}24x+27y=5,1\\x+1,5y=\dfrac{5,6}{22,4}=0,25\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Mg}=\dfrac{0,1\cdot24}{5,1}\cdot100\%=47,06\%\)
\(\%m_{Al}=100\%-47,06\%=52,94\%\)
b)\(\Sigma n_{HCl}=2x+3y=2\cdot0,1+3\cdot0,1=0,5mol\)
\(V=\dfrac{n}{C_M}=\dfrac{0,5}{2}=0,25l=250ml\)
\(n_{Mg}=\dfrac{6}{24}=0,25\left(mol\right)\\ PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
0,25 ---> 0,5 ---> 0,25 ---> 0,25
\(V_{H_2}=0,25.22,4=5,6\left(l\right)\\ m_{MgCl_2}=0,25.95=23,75\left(g\right)\\ m_{HCl}=0,5.36,5=18,25\left(g\right)\\ m_{ddHCl}=\dfrac{18,25}{18,25\%}=100\left(g\right)\\ m_{H_2}=0,25.2=0,5\left(g\right)\\ m_{dd}=100+6-0,5=105,5\left(g\right)\\ C\%_{MgCl_2}=\dfrac{23,75}{105,5}=22,51\%\)
Ta có: nHCl=0,7.0,1=0,07 mol
nNaOH=0,05.0,2=0,01 mol
NaOH + HCl \(\rightarrow\) NaCl + H2O\(\rightarrow\)nHCl phản ứng=0,07-0,01=0,06 mol
CaCO3 + 2HCl\(\rightarrow\) CaCl2 + CO2 + H2O
Na2CO3 + 2HCl\(\rightarrow\)2NaCl + CO2 + H2O
\(\rightarrow\) nCO2=1/2nHCl=0,03 mol\(\rightarrow\) V =0,03.22,4=0,672 lít
Gọi số mol CaCO3 là x; Na2CO3 là y \(\rightarrow\)100x +106y=3,12
\(\rightarrow\)nCO2=x+y=0,03
\(\rightarrow\) x=0,01; y=0,02
Muối thu được là CaCl2 0,01 mol và NaCl 0,04 mol
\(\rightarrow\)m muối=3,45 gam