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a) n H2 = 15,68/22,4 = 0,7(mol)
$Zn + 2HCl \to ZnCl_2 + H_2$
$Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : nHCl = 2n H2 = 1,4(mol)
=> CM HCl = 1,4/2 = 0,7M
b) n Zn = a(mol) ; n Fe = b(mol) => 65a + 56b = 43,7(1)
n H2 = a + b = 0,7(2)
Từ (1)(2) suy ra a = 0,5 ; b = 0,2
Suy ra:
m Zn = 0,5.65 = 32,5 gam
m Fe = 0,2.56 = 11,2 gam
a) n H2 = 15,68/22,4 = 0,7(mol)
Zn+2HCl→ZnCl2+H2Zn+2HCl→ZnCl2+H2
Fe+2HCl→FeCl2+H2Fe+2HCl→FeCl2+H2
Theo PTHH : nHCl = 2n H2 = 1,4(mol)
=> CM HCl = 1,4/2 = 0,7M
b) n Zn = a(mol) ; n Fe = b(mol) => 65a + 56b = 43,7(1)
n H2 = a + b = 0,7(2)
Từ (1)(2) suy ra a = 0,5 ; b = 0,2
Suy ra:
m Zn = 0,5.65 = 32,5 gam
m Fe = 0,2.56 = 11,2 gam
\(n_{Zn}=\dfrac{3,25}{65}=0,05\left(mol\right)\)
\(m_{ct}=\dfrac{14,6.50}{100}=7,3\left(g\right)\)
\(n_{HCl}=\dfrac{7,3}{36,5}=0,2\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,05 0,2 0,05 0,05
a) Lập tỉ số so sánh : \(\dfrac{0,05}{1}< \dfrac{0,2}{2}\)
⇒ Zn phản ứng hết , HCl dư
⇒ Tính toán dựa vào số mol của Zn
\(n_{H2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
b) \(n_{ZnCl2}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,05.136=6,8\left(g\right)\)
\(n_{HCl\left(dư\right)}=0,2-\left(0,5.2\right)=0,1\left(mol\right)\)
⇒ \(m_{HCl\left(dư\right)}=01,.36,5=3,65\left(g\right)\)
c) \(m_{ddspu}=3,25+50-\left(0,05.2\right)=53,15\left(g\right)\)
\(C_{ZnCl2}=\dfrac{6,8.100}{53,15}=12,8\)0/0
\(C_{HCl\left(dư\right)}=\dfrac{3,65.100}{53,15}=6,88\)0/0
Chúc bạn học tốt
a)
$n_{Zn} = \dfraac{3,25}{65} = 0,05(mol) ; n_{HCl} = \dfrac{50.14,6\%}{36,5} = 0,2(mol)$
$Zn +2 HCl \to ZnCl_2 + H_2$
$n_{Zn} : 1 < n_{HCl} : 2$ nên HCl dư
$n_{H_2} = n_{Zn} = 0,05(mol)$
$V_{H_2} = 0,05.22,4 = 1,12(lít)$
b)
$n_{ZnCl_2} = n_{Zn} = 0,05 \Rightarrow m_{ZnCl_2} = 0,05.136 = 6,8(gam_$
$n_{HCl\ pư} = 2n_{Zn} = 0,1(mol) \Rightarrow m_{HCl\ dư} = (0,2 - 0,1).36,5 = 3,65(gam)$
c)
$m_{dd\ sau\ pư} = 3,25 + 50 - 0,05.2 = 53,15(gam)$
d)
$C\%_{ZnCl_2} = \dfrac{6,8}{53,15}.100\%= 12,8\%$
$C\%_{HCl} = \dfrac{3,65}{53,15}.100\% = 6,87\%$
a) Gọi x,y lần lượt là số mol của Zn, Fe
nH2 = 15,68\22,4=0,7 mol
Pt: Zn + 2HCl --> ZnCl2 + H2
.....x......................................x
.....Fe + 2HCl --> FeCl2 + H2
.....y......................................y
Ta có hệ pt:
{65x+56y=43,7
x+y=0,7⇔{x=0,5,y=0,2
mZn = 0,5 . 65 = 32,5 (g)
mFe = 0,2 . 56 = 11,2 (g)
b) Pt: Fe3O4 + 4H2 --to--> 3Fe + 4H2O
.......................0,7 mol----> 0,525 mol
nFe3O4 = 46,4\232=0,2 mol
Xét tỉ lệ mol giữa Fe3O4 và H2:
0,2\1>0,7\4
Vậy Fe3O4 dư
mFe thu được = 0,525 . 56 = 29,4 (g)
b2
Giả sự đều sinh ra 1 mol H2
2Al+3H2SO4--->Al2(SO4)3+3H2
2/3_______________________1
=>a=2/3.27=18
Zn+H2SO4--->ZnSO4+H2
1___________________1
=>b=65
===> a/b=18/65
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,1}=2\left(M\right)\)
\(n_{Fe}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Fe}=0,1.56=5,6\left(g\right)\)
\(\Rightarrow m_{Cu}=20-5,6=14,4\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{5,6}{20}.100\%=28\%\\\%m_{Cu}=72\%\end{matrix}\right.\)
Gọi $n_{Fe} = a(mol), n_{Zn} = b(mol) , n_{Al} = c(mol) \Rightarrow 56a + 65b + 27c = 20,4(1)$
$Fe + H_2SO_4 \to FeSO_4 + H_2$
$Zn + H_2SO_4 \to ZnSO_4 + H_2$
$2Al +3 H_2SO_4 \to Al_2(SO_4)_3 +3 H_2$
Theo PTHH : $n_{H_2} = a + b + 1,5c = \dfrac{10,08}{22,4} = 0,45(mol)(2)$
Mặt khác :
$2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3$
$Zn + Cl_2 \xrightarrow{t^o} ZnCl_2$
$2Al + 3Cl_2 \xrightarrow{t^o} 2AlCl_3$
Theo PTHH : $n_{Cl_2} = 1,5n_{Fe} + n_{Zn} + 1,5n_{Al}$
Suy ra : \dfrac{1,5a + b + 1,5c}{a + b + c} = \dfrac{0,275}{0,2}(3)$
Từ (1)(2)(3) suy ra : a = 0,2 ; b = 0,1 ; c = 0,1
$\%m_{Fe} = \dfrac{0,2.56}{20,4}.100\% = 54,9\%$
$\%m_{Zn} = \dfrac{0,1.65}{20,4}.100\% = 31,9\%$
$\%m_{Al} = 100\% - 54,9\% - 31,9\% = 13,2\%$
a) Mg + 2HCl -> MgCl2 + H2
Al + 3HCl -> AlCl3 + 3/2H2
b) Gọi a, b lần lượt là số mol Mg, Al.
nH2 = 5,6/22,4 = 0,25 (mol)
Mg + 2HCl -> MgCl2 + H2
a 2a a a
Al + 3HCl -> AlCl3 + 3/2H2
b 3b b 3/2b
Ta có hệ pt:
mhh = 24a + 27b = 5,1 (g)
nH2 = a + 3/2b = 0,25 (mol)
=> a = 0,1 (mol)
b = 0,1 (mol)
200 ml = 0,2 l
nHCl = 2a + 3b = 0,2 + 0,3 = 0,5 (mol)
=> CM ddHCl = 0,5/0,2 = 2,5 (M)
%mMg = 24a/5,1*100% = 2,4/5,1*100% = 47,06%
%mAl = 100%-47,06% = 52,94%
\(a,m_{rắn}=m_{Cu}=2,7\left(g\right)\\ \Rightarrow m_{\left(Zn,Fe\right)}=12-2,7=9,3\left(g\right)\\ n_{H_2}=0,15\left(mol\right),n_{axit}=2.0,2=0,4\left(mol\right)\\ Đặt:n_{Zn}=a\left(mol\right);n_{Fe}=b\left(mol\right)\left(a,b>0\right)\\ PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ Vì:\dfrac{0,15}{1}< \dfrac{0,4}{1}\Rightarrow axit.dư\\ \Rightarrow\left\{{}\begin{matrix}65+56b=9,3\\a+b=\dfrac{3,36}{22,4}=0,15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,05\end{matrix}\right.\\ \Rightarrow\%m_{Cu}=\dfrac{2,7}{12}.100=22,5\%\\ \%m_{Zn}=\dfrac{0,1.65}{12}.100\approx54,167\%\\ \%m_{Fe}=\dfrac{0,05.56}{12}.100\approx23,333\%\)
\(b,ddA:FeCl_2,ZnCl_2,H_2SO_4\left(dư\right)\\ m_{ddH_2SO_4}=200.1,14=228\left(g\right)\\ m_{ddA}=m_{\left(Zn,Fe\right)}+m_{ddH_2SO_4}-m_{H_2}=9,3+228-0,15.2=237\left(g\right)\)
\(C\%_{ddZnCl_2}=\dfrac{136.0,1}{237}.100\approx5,738\%\\ C\%_{ddFeCl_2}=\dfrac{127.0,05}{237}.100\approx2,679\%\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{\left(0,4-0,15\right).98}{237}.100\approx10,338\%\)
Đã sửa lần cuối lúc 20:45
Gọi \(\left\{{}\begin{matrix}n_{Zn}=a\left(mol\right)\\n_{Fe}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ m_{HCl}=200.27,375\%=54,75\left(g\right)\\ n_{HCl}=\dfrac{54,75}{36,5}=1,5\left(mol\right)\)
PTHH:
Zn + 2HCl ---> ZnCl2 + H2
a ----> 2a --------> a -----> a
Fe + 2HCl ---> FeCl2 + H2
b ---> 2b -------> b ------> b
Hệ pt \(\left\{{}\begin{matrix}65a+56b=43,7\\a+b=0,7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,5\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}m_{Zn}=0,5.65=32,5\left(g\right)\\m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)
\(m_{dd}=43,7+200-0,7.2=242,3\left(g\right)\\ \rightarrow\left\{{}\begin{matrix}C\%_{ZnCl_2}=\dfrac{0,5.136}{242,3}=28,06\%\\C\%_{FeCl_2}=\dfrac{0,2.127}{242,3}=10,48\%\\C\%_{HCl\left(dư\right)}=\dfrac{\left(1,5-0,5.2-0,2.2\right).36,5}{242,3}=1,51\%\end{matrix}\right.\)
\(n_{H_2}=\dfrac{15,68}{22,4}=0,7\left(mol\right)\\ pthh:\left\{{}\begin{matrix}Zn+H_2SO_4->ZnSO_4+H_2\\Fe+H_2SO_4->FeSO_{\text{ 4 }}+H_2\end{matrix}\right.\)
gọi số mol Zn là x , số mol Fe là y
=> 65x+56y=43,7
=> a+b=0,7
=>a=0,5 , b =0,2
=> \(m_{Zn}=0,5.65=32,5\\ m_{Fe}=43,7-32,5=11,2\left(G\right)\)