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a)
\(n_{MgCl_2}=\dfrac{38}{95}=0,4\left(mol\right)\)
\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: MgCO3 + 2HCl --> MgCl2 + CO2 + H2O
0,3<------0,6<------0,3<----0,3
MgO + 2HCl --> MgCl2 + H2O
0,1<---0,2<------0,1
=> \(\left\{{}\begin{matrix}m_{MgO}=0,1.40=4\left(g\right)\\m_{MgCO_3}=0,3.84=25,2\left(g\right)\end{matrix}\right.\)
b) \(m_{HCl}=\left(0,6+0,2\right).36,5=29,2\left(g\right)\)
=> \(m_{dd.HCl}=\dfrac{29,2.100}{20}=146\left(g\right)\)
Gọi \(\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=0,2\cdot4=0,8mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(x\) \(\rightarrow\) \(3x\) \(x\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(y\) \(\rightarrow\) \(2y\) \(y\)
\(\Rightarrow\left\{{}\begin{matrix}27x+65y=11,9\\3x+2y=0,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
a)\(\%m_{Al}=\dfrac{0,2\cdot27}{11,9}\cdot100\%=45,38\%\)
\(\%m_{Zn}=100\%-45,38\%=54,62\%\)
b)\(\Sigma n_{H_2}=\dfrac{3}{2}x+y=\dfrac{3}{2}\cdot0,2+0,1=0,4mol\)
\(V_{H_2}=0,4\cdot22.4=8,96l\)
Na2SO3, CaSO3, Ba(HCO3) + HCl→SO2
Na2S, FeS + HCl→H2S
Na2SO3, CaSO3, Ba(HCO3) + HCl→SO2
Na2S, FeS + HCl→H2S
Na2SO3, CaSO3, Ba(HCO3) + HCl→SO2
Na2S, FeS + HCl→H2S
Na2SO3, CaSO3, Ba(HCO3) + HCl→SO2
Na2S, FeS + HCl→H2S
Na2SO3, CaSO3, Ba(HCO3) + HCl→SO2
Na2S, FeS + HCl→H2S
Câu 1:
Gọi : nMg=a(mol); nMgO=b(mol) (a,b>0)
a) PTHH: Mg + 2 HCl -> MgCl2 + H2
a________2a_______a______a(mol)
MgO +2 HCl -> MgCl2 + H2O
b_____2b_______b___b(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}24a+40b=8,8\\22,4a=4,48\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
=> mMg=0,2.24=4,8(g)
=>%mMg= (4,8/8,8).100=54,545%
=> %mMgO= 45,455%
b) m(muối)=mMg2+ + mCl- = 0,3. 24 + 0,6.35,5=28,5(g)
c) V=VddHCl=(2a+2b)/2=0,3(l)=300(ml)
Câu 2:
Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\n_{HCl}=0,4\cdot2=0,8\left(mol\right)\end{matrix}\right.\)
PTHH: \(Ca+2HCl\rightarrow CaCl_2+H_2\uparrow\)
0,2____0,4_____0,2____0,2 (mol)
\(CaO+2HCl\rightarrow CaCl_2+H_2O\)
0,2____0,4______0,2____0,2 (mol)
Ta có: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)
\(n_{H_2} = \dfrac{6,72}{22,4} = 0,3(mol)\\ Mg + 2HCl \to MgCl_2 + H_2\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)
\(\left\{{}\begin{matrix}Mg:x\left(mol\right)\\Al:y\left(mol\right)\end{matrix}\right.\)→ \(\left\{{}\begin{matrix}24x+27y=6,3\\x+1,5y=0,3\end{matrix}\right.\)→ \(\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\)
Vậy :
\(\%m_{Mg} = \dfrac{0,15.24}{6,3}.100\% = 57,14\%\)
\(n_{H_2}=\dfrac{2,464}{22,4}=0,11mol\)
\(\left\{{}\begin{matrix}Al:x\left(mol\right)\\Fe:y\left(mol\right)\end{matrix}\right.\Rightarrow Muối\left\{{}\begin{matrix}Al_2\left(SO_4\right)_3\\FeSO_4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}BTe:3x+2y=2n_{H_2}=0,22\\\dfrac{x}{2}\cdot342+y\cdot152=14,44\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,04mol\\y=0,05mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,04\cdot27=1,08g\\m_{Fe}=0,05\cdot56=2,8g\end{matrix}\right.\)
\(Al_2\left(SO_4\right)_3+3BaCl_2\rightarrow2AlCl_3+3BaSO_4\downarrow\)
0,02 0,06
\(FeSO_4+BaCl_2\rightarrow BaSO_4\downarrow+FeCl_2\)
0,05 0,05
\(\Rightarrow\Sigma n_{\downarrow}=0,06+0,05=0,11\Rightarrow m_{BaSO_4}=x=25,63g\)
\(n_{hhk}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Gọi x là số mol Ca
y là số mol CaCO3
\(Ca+2HCl\rightarrow CaCl_2+H_2\)
x..........2x...........x.............x
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\)
y...................2y............y........................y
Ta có x+y =0,3
Mặt khác ta lại có x:y=1:1
=> x=y=0,15
=>m=0,15.40+0,15.100=21(g)
\(m_{ddHCl}=\dfrac{\left(0,3+0,3\right).36,5}{14,6\%}=150\left(g\right)\)
mdd sau phản ứng = m + mddHCl-mkhí = 21 +150 - (0,15.2 + 0,15.44) = 164,1 (g)
=> \(C\%_{CaCl_2}=\dfrac{\left(0,15+0,15\right).111}{164,1}.100=20,29\%\)
Các chất khí tác dụng với dung dịch HCl tạo khí SO2, Na2SO3, CaSO3, Ba(HSO3)2
Na2CO3+2HCl→2NaCl+H2O+CO2
x---------------------------------------------x
CaCO3+2HCl→CaCl2+H2O+CO2
y----------------------------------------y
=>Ta có :\(\left\{{}\begin{matrix}126x +120y=36,6\\x+y=0,3\end{matrix}\right.\)
=>x=0,1 mol , y=0,2 mol
=>m Na2SO3=0,1.126=12,6g