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\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ a,Zn+2HCl\to ZnCl_2+H_2\\ b,n_{ZnCl_2}=0,1(mol)\\ \Rightarrow m_{ZnCl_2}=0,1.136=13,6(g)\\ c,n_{Zn}=0,1(mol)\\ \Rightarrow \%_{Zn}=\dfrac{0,1.65}{20}.100\%=32,5\%\\ \Rightarrow \%_{Ag}=100\%-32,5\%=67,5\%\)
a) \(n_{H_2}=0,2\left(mol\right)\)
Bảo toàn nguyên tố H : \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
=> \(n_{Cl^-}=0,4\left(mol\right)\)
=> \(m_{muối}=m_{KL}+m_{Cl^-}=20+0,4.35,5=34,2\left(g\right)\)
a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
\(Cu+2H_2SO_{4\left(đ\right)}\underrightarrow{t^o}CuSO_4+SO_2+2H_2O\)
b, Ta có: \(n_{H_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)
\(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(n_{SO_2}=\dfrac{2,9748}{24,79}=0,12\left(mol\right)\)
\(n_{Cu}=n_{SO_2}=0,12\left(mol\right)\)
\(\Rightarrow m=m_{Zn}+m_{Cu}=0,1.65+0,12.64=14,18\left(g\right)\)
Có: \(n_{H_2SO_{4\left(đ\right)}}=2n_{SO_2}=0,24\left(mol\right)\Rightarrow x=m_{ddH_2SO_4\left(đ\right)}=\dfrac{0,24.98}{98\%}=24\left(g\right)\)
Đặt: \(\left\{{}\begin{matrix}x=n_{Fe}\left(mol\right)\\y=n_{Al}\left(mol\right)\end{matrix}\right.\)
\(\sum m_{hh}=11\left(g\right)\Rightarrow56x+27y=11\left(1\right)\)
\(PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ \left(mol\right)....x\rightarrow..2x........x......x\\ PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\\ \left(mol\right)....y\rightarrow..3y.........y......1,5y\)
\(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\ \Rightarrow x+1,5y=0,4\left(2\right)\)
\(\xrightarrow[\left(2\right)]{\left(1\right)}\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)
a) \(\%m_{Fe}=\dfrac{56.0,1}{11}=51\%\)
\(\rightarrow\%m_{Al}=49\%\)
b) \(\sum m_{ctHCl}=\left(2.0,1+3.0,2\right).36,5=29,2\left(g\right)\)
\(m_{ddHCl}=\dfrac{29,2.100\%}{10\%}=292\left(g\right)\)
c) \(m_{H_2\uparrow}=\left(1.0,1+1,5.0,2\right).2=0,8\left(g\right)\)
\(m_{ddsaupu}=m_{hh}+m_{ddHCl}-m_{H_2\uparrow}=11+292-0,8=302,2\left(g\right)\)
\(C\%_{FeCl_2}=\dfrac{0,1.127}{302,2}.100=4,2\%\\ C\%_{AlCl_3}=\dfrac{0,2.133,5}{302,2}.100=8,8\%\)
Đặt: {x=nFe(mol)y=nAl(mol){x=nFe(mol)y=nAl(mol)
∑mhh=11(g)⇒56x+27y=11(1)∑mhh=11(g)⇒56x+27y=11(1)
PTHH:Fe+2HCl→FeCl2+H2↑(mol)....x→..2x........x......xPTHH:2Al+6HCl→2AlCl3+3H2↑(mol)....y→..3y.........y......1,5yPTHH:Fe+2HCl→FeCl2+H2↑(mol)....x→..2x........x......xPTHH:2Al+6HCl→2AlCl3+3H2↑(mol)....y→..3y.........y......1,5y
nH2=8,9622,4=0,4(mol)⇒x+1,5y=0,4(2)nH2=8,9622,4=0,4(mol)⇒x+1,5y=0,4(2)
(1)−→(2){x=0,1y=0,2→(2)(1){x=0,1y=0,2
a) %mFe=56.0,111=51%%mFe=56.0,111=51%
→%mAl=49%→%mAl=49%
b) ∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)∑mctHCl=(2.0,1+3.0,2).36,5=29,2(g)
mddHCl=29,2.100%10%=292(g)mddHCl=29,2.100%10%=292(g)
c) mH2↑=(1.0,1+1,5.0,2).2=0,8(g)mH2↑=(1.0,1+1,5.0,2).2=0,8(g)
mddsaupu=mhh+mddHCl−mH2↑=11+292−0,8=302,2(g)mddsaupu=mhh+mddHCl−mH2↑=11+292−0,8=302,2(g)
\(Mg + 2HCl \rightarrow MgCl_2 + H_2\)
\(Zn + 2HCl \rightarrow ZnCl_2 + H_2\)
\(2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2\)
\(n_{H_2}= \dfrac{10,08}{22,4}= 0, 45 mol\)
Theo PTHH:
\(n_{-Cl}= 2n_{H_2}= 0,9 mol\) ( gốc Cl ở muối nhé)
\(m_{muối}= m_{kim loại} + m_{-Cl} \Rightarrow 56,6=a + 0,9 . 35,5 \Rightarrow a=24,65g\)
1/ nH2 = 0,39 mol; nHCl = 0,5 mol; nH2SO4 = 0,14 mol
nH+= 0,5 + 0,14.2 = 0,78 = 2nH2
=> axit phản ứng vừa đủ
Bảo toàn khối lượng: mkim loại + mHCl + mH2SO4 = mmuối khan + mH2
=> mmuối khan = 7,74 + 0,5.36,5 + 0,14.98 – 0,39.2 = 38,93 gam
2/ Đặt x, y là số mol Mg, Al
\(\left\{{}\begin{matrix}24x+27y=7,74\\x+\dfrac{3}{2}y=0,39\end{matrix}\right.\)
=> x=0,12 ; y=0,18
Để thu được kết tủa lớn nhất thì Al(OH)3 không bị tan trong NaOH
Dung dịch A : Mg2+ (0,12 mol) , Al3+ (0,18 mol)
\(Mg^{2+}+2OH^-\rightarrow Mg\left(OH\right)_2\)
\(Al^{3+}+3OH^-\rightarrow Al\left(OH\right)_3\)
=> \(n_{OH^-}=n_{NaOH}=0,12.2+0,18.3=0,78\left(mol\right)\)
=> \(V_{NaOH}=\dfrac{0,78}{2}=0,39\left(lít\right)\)