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a) $Zn+ 2HCl \to ZnCl_2 + H_2$
b) $n_{HCl} = \dfrac{250.7,3\%}{36,5} = 0,5(mol)$
$n_{Zn} = n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol)$
$m = 0,25.65 =16,25(gam) ; V_{H_2} = 0,25.22,4 = 5,6(lít)$
c)
$m_{dd\ sau\ pư} = 16,25 + 250 - 0,25.2 = 265,75(gam)$
$C\%_{ZnCl_2} = \dfrac{0,25.136}{265,75}.100\% = 12,8\%$
\(a/ 4Zn+2HCl \to ZnCl_2+H_2 \\ n_{HCl}=\frac{250.7,3\%}{36,5}=0,5(mol)\\ b/ \\ n_{Zn}=n_{H_2}=n_{ZnCl_2}=\frac{1}{2}.n_{HCl}=\frac{1}{2}.0,5=0,25(mol)\\ m_{Zn}=0,25.65=16,25(g)\\ V_{H_2}=0,25.22,4=5,6(l)\\ c/ \\ C\%_{ZnCl_2}=\frac{0,25.136}{16,25+250-0,25.2}.100=12,8\% \)
mHCl= 3,65%.400= 14,6(g) => nHCl=14,6/36,5=0,4(mol)
a) PTHH: Mg +2 HCl -> MgCl2 + H2
0,2__________0,4_____0,2____0,2(mol)
V(H2,đktc)=0,2.22,4=4,48(l)
b)mMg=0,2.24=4,8(g)
c) mMgCl2= 0,2.95=19(g)
mddMgCl2= 400+4,8 - 0,2.2= 404,4(g)
=> C%ddMgCl2= (19/404,4).100=4,698%
400ml = 0,4l
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,1 0,1 0,1
a) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
\(n_{ZnCl2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(C_{M_{ZnCl2}}=\dfrac{0,1}{0,4}=0,25\left(M\right)\)
Chúc bạn học tốt
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ MgO+2HCl\rightarrow MgCl_2+H_2O\\ b.n_{H_2}=n_{Mg}=0,3\left(mol\right)\\ \Rightarrow\%m_{Mg}=\dfrac{0,3.24}{15,6}.100=48,15\%;\%m_{MgO}=53,85\%\)
a) PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\) (1)
\(ZnO+H_2SO_4\rightarrow ZnSO_4+H_2O\) (2)
b) Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\\Sigma n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow n_{Zn}=n_{ZnO}=n_{H_2SO_4\left(1\right)}=n_{H_2SO_4\left(2\right)}=0,1\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,1\cdot65=6,5\left(g\right)\\m_{ZnO}=0,1\cdot81=8,1\left(g\right)\end{matrix}\right.\)
c) Theo các PTHH: \(\left\{{}\begin{matrix}n_{Zn}=n_{ZnSO_4\left(1\right)}=0,1mol\\n_{ZnO}=n_{ZnSO_4\left(2\right)}=0,1mol\end{matrix}\right.\)
\(\Rightarrow\Sigma n_{ZnSO_4}=0,2mol\) \(\Rightarrow C_{M_{ZnSO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\)
(Coi như thể tích dd thay đổi không đáng kể)
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)
Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư
\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)
\(a)n_{Fe}=\dfrac{5,6}{56}=0,1mol\\ n_{Mg}=\dfrac{4,8}{24}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)
0,1 0,2 0,1 0,1
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,2 0,4 0,2 0,2
\(V_{H_2}=\left(0,1+0,2\right).22,4=6,72l\\ b)V_{ddHCl}=\dfrac{0,2+0,4}{2}=0,3l\\ c)m_{muối}=0,1.127+95.0,2=31,7g\)
a)
Zn+2HCl→ZnCl2+H2
b)
nHCl=250.7,3%/36,5=0,5(mol)
nZn=nH2=12nHCl=0,25(mol)
m=0,25.65=16,25(gam);VH2=0,25.22,4=5,6(lít)
c)
mdd sau pư=16,25+250−0,25.2=265,75(gam)
C%ZnCl2=0,25.136/265,75.100%=12,8%