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\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05 0,05
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,2 0,4 0,2
a) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(m_{Mg}=0,05.24=1,2\left(g\right)\)
\(m_{MgO}=9,2-1,2=8\left(g\right)\)
0/0Mg = \(\dfrac{1,2.100}{9,2}=13,04\)0/0
0/0MgO = \(\dfrac{8.100}{9,2}=86,96\)0/0
b) Có : \(m_{MgO}=8\left(g\right)\)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)
\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)
c) \(n_{MgCl2\left(tổng\right)}=0,05+0,2=0,25\left(mol\right)\)
⇒ \(m_{MgCl2}=0,25.95=23,75\left(g\right)\)
\(m_{ddspu}=9,2+125-\left(0,05.2\right)=134,1\left(g\right)\)
\(C_{MgCl2}=\dfrac{23,75.100}{134,1}=17,71\)0/0
Chúc bạn học tốt
a)\(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,05 0,1 0,05 0,05
PTHH: MgO + 2HCl → MgCl2 + H2O
Mol: 0,2 0,4 0,2
\(\%m_{Mg}=\dfrac{0,05.24.100\%}{9,2}=13,04\%;\%m_{MgO}=100-13,04=86,96\%\)
\(n_{MgO}=\dfrac{9,2-0,05.24}{40}=0,2\left(mol\right)\)
b,\(m_{HCl}=\left(0,1+0,4\right).36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)
c,mdd sau pứ = 9,2+125-0,05.2 = 134,1 (g)
\(C\%_{ddMgCl_2}=\dfrac{\left(0,05+0,2\right).95.100\%}{134,1}=17,71\%\)
\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)
\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)
\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)
a) 2Al + 6HCl --> 2AlCl3 + 3H2
Fe + 2HCl --> FeCl2 + H2
b) Gọi số mol Al, Fe lần lượt là a,b
=> 27a + 56b = 13,9
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
2Al + 6HCl --> 2AlCl3 + 3H2
a----->3a--------->a------->1,5a______(mol)
Fe + 2HCl --> FeCl2 + H2
b------>2b-------->b----->b__________(mol)
=> 1,5a + b = 0,35
=> \(\left\{{}\begin{matrix}a=0,1=>m_{Al}=0,1.27=2,7\left(g\right)\\b=0,2=>m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)
c) nHCl = 3a + 2b = 0,7 (mol)
=> mHCl = 0,7.36,5 = 25,55(g)
=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)
\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)
\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)
PTHH: \(NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2\uparrow\)
a_____a_______a_____a_____a (mol)
\(Na_2CO_3+2HCl\rightarrow2NaCl+H_2O+CO_2\uparrow\)
b_____2b_______2b____b_____b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}84a+106b=38\\a+b=\dfrac{8,96}{22,4}=0,4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,2\\b=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{NaHCO_3}=\dfrac{0,2\cdot84}{38}\cdot100\%\approx44,21\%\\\%m_{Na_2CO_3}=55,79\%\end{matrix}\right.\)
Mặt khác: \(n_{NaCl}=0,6\left(mol\right)\) \(\Rightarrow m_{NaCl}=0,6\cdot58,5=35,1\left(g\right)\)
a) Đặt: nNa2CO3=x(mol); nNaHCO3=y(mol) (x,y>0)
PTHH: Na2CO3 + 2 HCl -> 2 NaCl + CO2+ H2O
x_______________2x____2x_______x(mol)
NaHCO3 + HCl -> NaCl + H2O + CO2
y__________y____y_____________y(mol)
Ta có hpt:
\(\left\{{}\begin{matrix}106x+84y=38\\x+y=0,4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)
b) mNaHCO3= 0,2. 84= 16,8(g)
=>%mNaHCO3= (16,8/38).100=44,211%
c) m(muối thu)= mNaCl(tổng)= (2x+y).58,5=0,6.58,5=35,1(g)
\(PTHH:Na_2SO_3+2HCl\rightarrow2NaCl+H_2O+SO_2\uparrow\\ K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\\ n_{SO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}+n_{K_2SO_3}=0,2\\126n_{Na_2SO_3}+158n_{K_2SO_3}=28,4\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}n_{Na_2SO_3}=0,1\left(mol\right)\\n_{K_2SO_3}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\%_{m_{Na_2SO_3}}=\dfrac{0,1\cdot126}{28,4}\cdot100\%\approx44\%\\ \Rightarrow\%_{m_{K_2SO_3}}=100\%-44\%=56\%\)
\(n_{HCl}=0,1\cdot2+0,1\cdot2=0,4\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,4\cdot36,5=14,6\left(g\right)\\ \Rightarrow C\%_{HCl}=\dfrac{14,6}{200}\cdot100\%=7,3\%CC\)
\(n_{MnO_2}=\dfrac{3,48}{87}=0,04(mol)\\ n_{Cl_2}=\dfrac{0,672}{22,4}=0,03(mol)\\ a,MnO_2+4HCl\xrightarrow{t^o}MnCl_2+Cl_2+2H_2O\\ \Rightarrow n_{Cl_2(p/ứ)}=0,04(mol)\\ \Rightarrow V_{Cl_2(p/ứ)}=0,04.22,4=0,896(l)\\ \Rightarrow H\%=\dfrac{0,672}{0,896}.100\%=75\%\)
\(b,n_{Cu}=\dfrac{0,64}{64}=0,01(mol)\\ 2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3(1)\\ Cu+Cl_2\xrightarrow{t^o}CuCl_2(2)\\ \Rightarrow n_{Cl_2(2)}=n_{Cu}=0,01(mol)\\ \Rightarrow n_{Cl_2(1)}=0,03-0,01=0,02(mol)\\ \Rightarrow n_{Fe}=\dfrac{1}{75}(mol) \Rightarrow m_{Fe}=\dfrac{1}{75}.56=0,75(g)\\ \Rightarrow m_{hh}=0,75+0,64=1,39(g)\)
\(c,FeCl_3+3AgNO_3\to Fe(NO_3)_3+3AgCl\downarrow\\ CuCl_2+2AgNO_3\to Cu(NO_3)_2+2AgCl\downarrow\\ \Rightarrow \Sigma n_{AgCl}=3n_{FeCl_3}+2n_{CuCl_2}=0,04+0,01=0,05(mol)\\ \Rightarrow m_{\downarrow}=\Sigma m_{AgCl}=0,05.143,5=7,175(g)\\ d,FeCl_3+3NaOH\to Fe(OH)_3\downarrow+3NaCl\\ CuCl_2+2NaOH\to Cu(OH)_2\downarrow+2NaCl\\ \Rightarrow \Sigma n_{naOH}=3n_{FeCl_3}+2n_{CuCl_2}=0,05(mol)\\ \Rightarrow V_{dd_{NaOH}}=\dfrac{0,05}{2}=0,025(mol)\\ \Rightarrow m_{dd_{NaOH}}=0,025.1,12=0,028(g)\)
a)
PTHH: 2A + 2nHCl --> 2ACln + nH2
2B + 2mHCl --> 2BClm + mH2
Gọi số mol H2 là a (mol)
=> nHCl = 2a (mol)
Theo ĐLBTKL: mkim loại + mHCl = mmuối + mH2
=> 8,9 + 36,5.2a = 23,1 + 2a
=> a = 0,2 (mol)
=> VH2 = 0,2.22,4 = 4,48 (l)
b)
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
=> \(n_{H_2\left(tăng\right)}=0,25-0,2=0,05\left(mol\right)\)
PTHH: 2B + 2mHCl --> 2BClm + mH2
\(\dfrac{0,1}{m}\)<------------\(\dfrac{0,1}{m}\)<---0,05
Khối lượng rắn sau pư tăng lên do có thêm BClm sinh ra
=> \(m_{BCl_m}=\dfrac{0,1}{m}\left(M_B+35,5m\right)=27,85-23,1=4,75\left(g\right)\)
=> MB = 12m (g/mol)
Xét m = 2 thỏa mãn => MB = 24 (g/mol) => B là Mg
\(n_{Mg\left(thêm\right)}=\dfrac{0,1}{m}=\dfrac{0,1}{2}=0,05\left(mol\right)\)
=> \(n_{Mg\left(bđ\right)}=0,1\left(mol\right)\)
=> \(m_A=8,9-0,1.24=6,5\left(g\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
0,1-------------------->0,1
2A + 2nHCl --> 2ACln + nH2
\(\dfrac{0,2}{n}\)<-------------------0,1
=> \(M_A=\dfrac{6,5}{\dfrac{0,2}{n}}=32,5n\left(g/mol\right)\)
Xét n = 2 thỏa mãn => MA = 65 (g/mol)
=> A là Zn
nH2=0,84(mol)
PTHH: Fe + 2 HCl -> FeCl2 + H2
0,84__________________0,84(mol)
=>mFe=0,84.56=47,04(g)
=> mCaCO3=48,8-47,04=1,76(g)
b) %mCaCO3=(1,76/48,8).100=3,607%
=>%mFe= 96,393%
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
\(CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2\uparrow\)
Ta có: \(n_{Fe}=n_{H_2}=\dfrac{18,816}{22,4}=0,84\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,84\cdot56=47,04\left(g\right)\) \(\Rightarrow m_{CaCO_3}=1,76\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CaCO_3}=\dfrac{1,76}{48,8}\cdot100\%\approx3,61\%\\\%m_{Fe}=96,39\%\end{matrix}\right.\)
- Gọi x,y lần lượt là khối lượng của CaCO3 và MgCO trong hh (x,y >0) (g)
nCaCO3= x(mol)=> mCaCO3= 100x(g)
nMgCO3= y(mol) => mMgCO3= 84y(g)
=> mCaCO3+ mMgCO3= mhh
<=> 100x+ 84y= 2,84 (a)
PTHH: (1) CaCO3 + 2 HCl -> CaCl2 + H2O + CO2
(2) MgCO3 + 2 HCl -> MgCl2 + H2O + CO2
Ta có: nCO2(tổng)= (672:1000):22,4= 0,03(mol)
nCO2(1)= nCaCO3= x(mol); nCO2(2)= nMgCO3= y(mol)
=> nCO2(1) + nCO2(2)= nCO2(tổng)
<=> x+y= 0,03 (b)
Từ (a),(b) ta có hpt:
\(\left\{{}\begin{matrix}100x+84y=2,84\\x+y=0,03\end{matrix}\right.\)
Giaỉ hpt: x= 0,02 ; y= 0,01
=> mCaCO3= 100x= 100. 0,02= 2(g)
=> %mCaCO3= (2/ 2,84).100 \(\approx\) 70,423%
=> %mMgCO3 \(\approx\)100% - 70,423%= 29,577%