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a) $n_{H_2SO_4} = \dfrac{44,1}{98} = 0,45(mol)$
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
Theo PTHH :
$n_{Al} = \dfrac{2}{3}n_{H_2SO_4} = 0,3(mol)$
$m_{Al} = 0,3.27 = 8,1(gam)$
b) $n_{H_2} = n_{H_2SO_4} = 0,45(mol)$
$\Rightarrow V_{H_2} = 0,45.22,4 =1 0,08(lít)$
c)
Cách 1 : $n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = 0,15(mol)$
$\Rightarrow m_{Al_2(SO_4)_3} = 0,15.342 = 51,3(gam)$
Cách 2 : Bảo toàn khối lượng, $m_{Al_2(SO_4)_3} = 8,1 + 44,1 - 0,45.2 = 51,3(gam)$
Ta có: \(n_{H_2SO_4}=\dfrac{58,8}{98}=0,6\left(mol\right)\)
\(PTHH:2Al+3H_2SO_4--->Al_2\left(SO_4\right)_3+3H_2\uparrow\)
0,4 <--- 0,6 -----------> 0,2 --> 0,6
\(\Rightarrow\left\{{}\begin{matrix}m_{Al}=0,4.27=10,8\left(g\right)\\m_{Al_2\left(SO_4\right)_3}=0,2.342=68,4\left(g\right)\\V_{H_2}=0,6.22,4=13,44\left(lít\right)\end{matrix}\right.\)
a, Gọi \(\left\{{}\begin{matrix}n_{Fe}=a\left(mol\right)\\n_{Al}=b\left(mol\right)\end{matrix}\right.\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH:
Fe + 2HCl ---> FeCl2 + H2
a--->2a------------------>a
2Al + 6HCl ---> 2AlCl3 + 3H2
b---->3b-------------------->1,5b
=> \(\left\{{}\begin{matrix}56a+27b=16,6\\a+1,5b=0,5\end{matrix}\right.\Leftrightarrow a=b=0,2\left(mol\right)\)
=> \(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=0,2.27=5,4\left(g\right)\end{matrix}\right.\)
b) \(C\%_{HCl}=\dfrac{\left(0,2.2+0,2.3\right).36,5}{300}.100\%=12,167\%\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
gọi nFe : a , nAl: b (a,b>0) => 56a + 27b = 16,6 (g)
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a a
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
b \(\dfrac{3b}{2}\)
=> \(a+\dfrac{3b}{2}=0,5\)
ta có hệ pt
\(\left\{{}\begin{matrix}56a+27b=16,6\\a+\dfrac{3b}{2}=0,5\end{matrix}\right.\)
=> a= 0,2 , b = 0,2
\(\left\{{}\begin{matrix}m_{Fe}=0,2.56=11,2\left(g\right)\\m_{Al}=16,6-11,2=5,4\left(g\right)\end{matrix}\right.\)
\(pthh:Fe+2HCl\rightarrow FeCl_2+H_2\)
0,2 0,4
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,2 0,6
=> \(m_{HCl}=\left(0,4+0,6\right).36,5=36,5\left(g\right)\)
=> \(C\%=\dfrac{36,5}{200}.100\%=18,25\%\)
Giải:
Số mol của H2 là:
nH2 = V/22,4 = 19,6/22,4 = 0,875 (mol)
Gọi nMg = x (mol) và nAl = y (mol)
PTHH: Mg + 2HCl -> MgCl2 + H2↑
---------x--------------------------x--
PTHH: 2Al + 6HCl -> 2AlCl3 + 3H2↑
----------y-----------------------------\(\dfrac{3}{2}y\)--
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}m_{Mg}=m_{Al}\\n_{H_2}=0,875\left(mol\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}24x-27y=0\\x+\dfrac{3}{2}y=0,875\left(mol\right)\end{matrix}\right.\)
Giải hệ phương trình, ta được:
\(\left\{{}\begin{matrix}x=0,375\left(mol\right)\\y=\dfrac{1}{3}\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=n.M=0,375.24=9\left(g\right)\\m_{Al}=n.M=\dfrac{1}{3}.27=9\left(g\right)\end{matrix}\right.\)
Khối lượng hỗn hợp kim loại là:
\(m_{Mg}+m_{Al}=9+9=18\left(g\right)\)
Vậy ...
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{19,6}{22,4}=0,875\left(mol\right)\)
Đặt \(m_{Mg}=m_{Al}=a\left(g\right)\left(a>0\right)\)
\(\Rightarrow n_{Mg}=\dfrac{m}{M}=\dfrac{a}{24}\left(mol\right)\\ n_{Al}=\dfrac{m}{M}=\dfrac{a}{27}\left(mol\right)\)
\(pthh:Mg+HCl\rightarrow MgCl_2+H_2\left(1\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\left(2\right)\)
Theo \(pthh\left(1\right):n_{H_2\left(1\right)}=n_{Al}=\dfrac{a}{24}\left(mol\right)\)
Theo \(pthh\left(1\right):n_{H_2\left(1\right)}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}\cdot\dfrac{a}{27}=\dfrac{a}{18}\left(mol\right)\)
\(\Rightarrow\Sigma n_{H_2}=n_{H_2\left(1\right)}+n_{H_2\left(2\right)}\\ \Rightarrow0,875=\dfrac{a}{24}+\dfrac{a}{18}\\ \Rightarrow\dfrac{7}{72}a=0,875=9\left(T/m\right)\)
\(m_{h^2}=2a=2\cdot9=18\left(g\right)\)
a) Fe + 2HCl --> FeCl2 + H2
b) \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
_____0,1--------------->0,1---->0,1
=> mFeCl2 = 0,1.127 = 12,7(g)
c) VH2 = 0,1.22,4 = 2,24(l)
\(n_{Fe}=\dfrac{5,6}{56}=0,1(mol)\\ a,Fe+2HCl\to FeCl_2+H_2\\ \Rightarrow n_{FeCl_2}=n_{H_2}=0,1(mol)\\ a,m_{FeCl_2}=0,1.127=12,7(g)\\ b,V_{H_2}=0,1.22,4=2,24(l)\)
Bài 1 :
Giả sử : hỗn hợp có 1 mol
\(n_{H_2}=a\left(mol\right),n_{O_2}=1-a\left(mol\right)\)
\(\overline{M_X}=0.3276\cdot29=9.5\left(\dfrac{g}{mol}\right)\)
\(\Rightarrow m_X=2a+32\cdot\left(1-a\right)=9.5\left(g\right)\)
\(\Rightarrow a=0.75\)
Cách 1 :
\(\%H_2=\dfrac{0.75}{1}\cdot100\%=75\%\)
\(\%O_2=100-75=25\%\)
Cách 2 em tính theo thể tích nhé !
