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a) Sửa đề: dd H2SO4 9,8%
Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)
Bảo toàn nguyên tố: \(n_{H_2SO_4}=n_{H_2}=0,35\left(mol\right)\) \(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,35\cdot98}{9,8\%}=350\left(g\right)\)
\(\Rightarrow m_{dd}=m_{KL}+m_{H_2SO_4}-m_{H_2}=361,6\left(g\right)\)
b) Tương tự câu a
\(a) Mg + H_2SO_4 \to MgSO_4 + H_2\\ n_{Mg} = n_{H_2} = \dfrac{1,12}{22,4} =0,05(mol)\\ m_{Mg} = 0,05.24 =1,2(gam)\\ m_{Cu} = 7,6 -1,2 = 6,4(gam)\\ b) n_{H_2SO_4} = n_{H_2} = 0,05(mol) \Rightarrow V_{dd\ H_2SO_4} = \dfrac{0,05}{0,5} =0,1(lít)\\ c) n_{MgSO_4} = n_{H_2} = 0,05(mol) \Rightarrow m_{MgSO_4} = 0,05.120 = 6(gam)\\ d) \text{Bảo toàn electron: } 2n_{Mg} + 2n_{Cu} = 2n_{SO_2}\\ \Rightarrow n_{SO_2} = 0,05 + \dfrac{6,4}{64} = 0,15(mol) \Rightarrow V_{SO_2} = 0,15.22,4 = 3,36(lít)\)
a.Mg + H2SO4 -> MgSO4 + H2
b.\(nH2=\dfrac{4.704}{22.4}=0.21mol\) = nMg
mMg = 0.21\(\times24=5.04g\)
\(\%mMg=\dfrac{5.04\times100}{25}=20.16\%\)
\(\%mAg=100-20.16=79.84\%\)
c.MgSO4 + 2KOH -> K2SO4 + Mg(OH)2
0.21 0.42
H2SO4 + 2KOH -> K2SO4 + H2O
0.04 0.08
\(nH2SO4=\dfrac{9.8\times250}{100\times98}=0.25mol\)
Mà nH2SO4 phản ứng = nH2 = 0.21 mol
\(\Rightarrow nH2SO4dư=0.25-0.21=0.04mol\)
=> nKOH = 0.42 + 0.08 = 0.5mol
\(\Rightarrow CM_{KOH}=\dfrac{0.5}{0.625}=0.8M\)
\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
x x
\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
y 3/2y
\(nH_2=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)
ta có hệ pt:
\(\left\{{}\begin{matrix}x+\dfrac{3}{2}y=0,4\\24x+27y=7,8\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
\(mMg=0,1.24=2,4\left(g\right)\)
\(mAl=0,2.27=5,4\left(g\right)\)
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: \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{0,2\cdot40}{0,2\cdot40+0,2\cdot56}\cdot100\%\approx41,67\%\\\%m_{CaO}=58,33\%\\m_{CaCl_2}=\left(0,2+0,2\right)\cdot111=44,4\left(g\right)\end{matrix}\right.\)
\(n_{H2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Pt : \(Fe+2HCl\rightarrow FeCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
\(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O|\)
1 6 2 3
0,2 1,2 0,4
\(n_{Fe}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{Fe}=0,2.56=11,2\left(g\right)\)
\(m_{Fe2O3}=27,2-11,2=16\left(g\right)\)
0/0Fe = \(\dfrac{11,2.100}{27,2}=41,18\)0/0
0/0Fe2O3 = \(\dfrac{16.100}{27,2}=58,82\)0/0
b) Có : \(m_{Fe2O3}=16\left(g\right)\)
\(n_{Fe2O3}=\dfrac{16}{160}=0,1\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,4+1,2=1,6\left(mol\right)\)
\(V_{HCl}=\dfrac{1,6}{2}=0,8\left(l\right)\)
c) \(n_{FeCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(n_{FeCl3}=\dfrac{1,2.2}{6}=0,4\left(mol\right)\)
\(C_{M_{FeCl2}}=\dfrac{0,2}{0,8}=0,25\left(M\right)\)
\(C_{M_{FeCl3}}=\dfrac{0,4}{0,8}=0,5\left(M\right)\)
Chúc bạn học tốt
a)\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: x 1,5x
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: y y
Ta có: \(\left\{{}\begin{matrix}27x+24y=5,1\\1,5x+y=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)
\(\%m_{Al}=\dfrac{0,1.27.100\%}{5,1}=52,94\%;\%m_{Mg}=100-52,94=47,06\%\)
b)
PTHH: 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
Mol: 0,1 0,15 0,05
PTHH: Mg + H2SO4 → MgSO4 + H2
Mol: 0,1 0,1 0,1
\(m_{ddH_2SO_4}=\dfrac{\left(0,1+0,15\right).98.100}{9,8}=250\left(g\right)\)
mdd sau pứ = 5,1+250-0,15.2 = 254,8(g)
\(C\%_{ddAl_2\left(SO_4\right)_3}=\dfrac{0,05.342.100\%}{254,8}=6,71\%\)
\(C\%_{ddMgSO_4}=\dfrac{0,1.120.100\%}{254,8}=4,71\%\)
