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\(n_{H_2}=\frac{6,72}{22,4}=0,3mol\)
PTHH:
\(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Theo PTHH: \(n_{HCl}=2.n_{H_2}=2.0,3=0,6mol\)
HCl lấy dư 10%
\(\rightarrow V_{HCl}=\frac{0,6}{1}\left(100\%+10\%\right)=0,66l\)
a)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Mg + 2HCl --> MgCl2 + H2
_____0,1<---0,2<-------0,1<---0,1
=> mHCl = 0,2.36,5 = 7,3 (g)
=> \(m_{ddHCl}=\dfrac{7,3.100}{7,3}=100\left(g\right)\)
mdd sau pư = 0,1.24 + 100 - 0,1.2 = 102,2 (g)
\(C\%\left(MgCl_2\right)=\dfrac{0,1.95}{102,2}.100\%=9,2955\%\)
b)
CTHH: AaOb
PTHH: \(A_aO_b+2bHCl->aACl_{\dfrac{2b}{a}}+bH_2O\)
____________0,2------->\(\dfrac{0,1a}{b}\)
=> \(\dfrac{0,1a}{b}\left(M_A+35,5.\dfrac{2b}{a}\right)=13,5\)
=> \(M_A=\dfrac{64b}{a}=\dfrac{2b}{a}.32\)
Nếu \(\dfrac{2b}{a}=1\) => MA = 32 (L)
Nếu \(\dfrac{2b}{a}=2\) => MA = 64(Cu)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b, \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
c, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Ta có: \(n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,2}{1}\), ta được CuO dư.
Theo PT: \(n_{Cu}=n_{H_2}=0,2\left(mol\right)\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2HCl\rightarrow MgCl_2+H_2\\ n_{HCl}=2.0,2=0,4\left(mol\right);n_{H_2}=n_{Mg}=0,2\left(mol\right)\\ b,m_{HCl}=0,4.36,5=14,6\left(g\right)\\ c,n_{CuO}=\dfrac{24}{80}=0,3\left(mol\right)\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Vì:\dfrac{0,2}{1}< \dfrac{0,3}{1}\Rightarrow CuOdư\\ n_{Cu}=n_{H_2}=0,2\left(mol\right)\\ m_{Cu}=0,2.64=12,8\left(g\right)\)
a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)
Ta có: \(n_{Mg}=\dfrac{7,2}{24}=0,3\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Mg}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
b, \(n_{Fe_2O_3}=\dfrac{48}{160}=0,3\left(mol\right)\)
PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,3}{3}\), ta được Fe2O3 dư.
Theo PT: \(n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,1\left(mol\right)\Rightarrow n_{Fe_2O_3\left(dư\right)}=0,3-0,1=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe_2O_3}=0,2.160=32\left(g\right)\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a. PTHH:
Mg + 2HCl ---> MgCl2 + H2 (1)
MgO + 2HCl ---> MgCl2 + H2O (2)
Theo PT(1): \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\)
=> \(m_{Mg}=0,1.24=2,4\left(g\right)\)
=> \(m_{MgO}=4,4-2,4=2\left(g\right)\)
b. Ta có: \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)
=> \(n_{hh}=0,05+0,1=0,15\left(mol\right)\)
Theo PT(1,2): \(n_{HCl}=2.n_{hh}=2.0,15=0,3\left(mol\right)\)
=> \(V_{dd_{HCl}}=\dfrac{0,3}{0,4}=0,75\left(lít\right)\)
a)
Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Mg}=b\left(mol\right)\end{matrix}\right.\)
=> 56a + 24b = 18,4 (1)
PTHH: Fe + 2HCl --> FeCl2 + H2
a-->2a------>a------>a
Mg + 2HCl --> MgCl2 + H2
b--->2b------->b------>b
=> \(a+b=\dfrac{11,2}{22,4}=0,5\) (2)
(1)(2) => a = 0,2 (mol); b = 0,3 (mol)
\(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,2.56}{18,4}.100\%=60,87\%\\\%m_{Mg}=\dfrac{0,3.24}{18,4}.100\%=39,13\%\end{matrix}\right.\)
b) \(n_{HCl\left(pư\right)}=2a+2b=1\left(mol\right)\)
=> \(n_{HCl\left(tt\right)}=\dfrac{1.125}{100}=1,25\left(mol\right)\)
=> mHCl(tt) = 1,25.36,5 = 45,625 (g)
=> \(a=\dfrac{45,625.100}{18,25}=250\left(g\right)\)
c)
mdd sau pư = 18,4 + 250 - 0,5.2 = 267,4 (g)
\(C\%_{FeCl_2}=\dfrac{0,2.127}{267,4}.100\%=9,5\%\)
\(C\%_{MgCl_2}=\dfrac{0,3.95}{267,4}.100\%=10,66\%\)
1)
$MgO + 2HCl to MgCl_2 + H_2O$
$Mg + 2HCl \to MgCl_2 + H_2$
2)
$n_{Mg} = n_{H_2} = \dfrac{2,24}{22,4} = 0,1(mol)$
$m_{Mg} = 0,1.24 = 2,4(gam)$
$m_{MgO} = 4,4 - 2,4 = 2(gam)$
3)
$n_{HCl} = 2n_{Mg} + 2n_{MgO} = 0,1.2 + \dfrac{2}{40}.2 = 0,3(mol)$
$V_{dd\ HCl} = \dfrac{0,3}{2} = 0,15(lít) = 150(ml)$
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b) Ta có: \(n_{HCl\left(p/ứ\right)}=2n_{Mg}=2\cdot\dfrac{7,2}{24}=0,6\left(mol\right)\)
\(\Rightarrow n_{HCl}=0,6\cdot110\%=0,66\left(mol\right)\) \(\Rightarrow m_{ddHCl}=\dfrac{0,66\cdot36,5}{7,3\%}=330\left(g\right)\)
c) PTHH: \(Fe_2O_3+3H_2\xrightarrow[]{t^o}2Fe+3H_2O\)
Theo PTHH: \(n_{Fe}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}n_{Mg}=0,2\left(mol\right)\) \(\Rightarrow m_{Fe}=0,2\cdot56=11,2\left(g\right)\)