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\(2Al + 6HCl \rightarrow 2AlCl_3 + 3H_2\)
\(n_{Al}= \dfrac{5,4}{27}= 0,2 mol\)
Theo PTHH:
\(n_{AlCl_3}= n_{Al}= 0,2 mol\)
\(\Rightarrow m_{AlCl_3}= 0,2 . 133,5=26,7 g\)
Theo PTHH:
\(n_{H_2}= \dfrac{3}{2} n_{Al}= 0,3 mol\)
\(\Rightarrow V= 0,3 . 22,4= 6,72 l\)
b)
Theo PTHH:
\(n_{HCl}= 3n_{Al}= 0,6 mol\)
\(\Rightarrow m_{HCl}= 0,6 . 36,5=21,9 g\)
\(\Rightarrow m_{dd HCl}= \dfrac{21,9 . 100}{15}= 146 g\) ( nếu ở tử là : 21,9 . 100% thì ở mẫu bạn chia cho 15% nhé)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,1(mol)\\ \Rightarrow \%_{Mg}=\dfrac{0,1.24}{6,4}.100\%=37,5\%\\ \Rightarrow \%_{MgO}=100\%-37,5\%=62,5\%\)
\(b,n_{MgO}=\dfrac{6,4-0,1.24}{40}=0,1(mol)\\ \Rightarrow n_{HCl}=2.0,1+2.0,1=0,4(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,4}{0,5}=0,8(l)\\ c,n_{MgCl_2}=0,1+0,1=0,2(mol)\\ \Rightarrow C_{M_{MgCl_2}}=\dfrac{0,2}{0,8}=0,25M\)
1.1. Al + NaOH + H2O ==> NaAlO2 + 3/2H2
nH2(1)=3,36/22,4=0.15(mol)
=> nAl(1)= nH2(1):3/2= 0.15:3/2= 0.1(mol)
2.Mg + 2HCl ==> MgCl2 + H2
3.2Al + 6HCl ==> 2AlCl3 + 3H2
4.Fe + 2HCl ==> FeCl2 + H2
=> \(n_{H_2\left(2,3,4\right)}=\) 10.08/22.4= 0.45(mol)
=> nH2(3)=0.1*3/2=0.15(mol)
MgCl2 + 2NaOH ==> Mg(OH)2 + 2NaCl
AlCl3 + 3NaOH ==> Al(OH)3 + 3NaCl
FeCl2 + 2NaOH ==> Fe(OH)2 + 2NaCl
PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
2a_____4a______2a____2a (mol)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a_____2a______a_____a (mol)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
b______3b______b______\(\dfrac{3}{2}\)b (mol)
Ta lập HPT: \(\left\{{}\begin{matrix}2a+a+\dfrac{3}{2}b=\dfrac{13,44}{22,4}=0,6\\24\cdot2a+56a+27b=15,8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{0,1\cdot56}{15,8}\cdot100\%=35,44\%\\\%m_{Al}=\dfrac{0,2\cdot27}{15,8}=34,18\%\\\%m_{Mg}=30,38\%\end{matrix}\right.\)
Theo các PTHH: \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{1,2\cdot36,5}{200}\cdot100\%=21,9\%\)
a) PTHH: \(Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+SO_2\uparrow\)
Ta có: \(n_{Na_2SO_3}=\dfrac{12,6}{126}=0,1\left(mol\right)=n_{SO_2}\) \(\Rightarrow V_{SO_2}=0,1\cdot22,4=2,24\left(l\right)\)
b) Ta có: \(\left\{{}\begin{matrix}n_{SO_2}=0,1\left(mol\right)\\n_{Ca\left(OH\right)_2}=1,4\cdot0,1=0,14\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Tạo muối trung hòa
PTHH: \(Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3\downarrow+H_2O\)
Ta có: \(\left\{{}\begin{matrix}n_{CaSO_3}=0,1\left(mol\right)=n_{Ns_2SO_4}\\n_{Ca\left(OH\right)_2\left(dư\right)}=0,04\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{CaSO_3}=0,1\cdot120=12\left(g\right)\\m_{Na_2SO_4}=0,1\cdot142=14,2\left(g\right)\\m_{Ca\left(OH\right)_2\left(dư\right)}=0,04\cdot74=2,96\left(g\right)\end{matrix}\right.\)
\(n_{Na_2SO_3}=0,1\left(mol\right)\\ PTHH:Na_2SO_3+H_2SO_4\rightarrow Na_2SO_4+H_2O+SO_2\)
(mol) 0,1 0,1 0,1 0,1
\(a.V_{SO_2}=0,1.22,4=2,24\left(l\right)\)
\(b.n_{Ca\left(OH\right)_2}=0,14\left(mol\right)\)
Do \(\dfrac{n_{OH}}{n_{SO_2}}=\dfrac{0,28}{0,1}=2.8>2\rightarrow\) Tạo muối trung hòa và Ca(OH)2 dư 0,04(mol)
\(PTHH:Ca\left(OH\right)_2+SO_2\rightarrow CaSO_3+H_2O\)
(mol) 0,1 0,1 0,1 0,1
\(m_{Ca\left(OH\right)_2\left(du\right)}=0,04.74=2,96\left(g\right)\\ m_{CaSO_3}=12\left(g\right)\\ m_{H_2O}=0,1.18=1,8\left(g\right)\)
CH4+2O2-to>CO2+2H2O
x---------2x-------x
C2H4+3O2-to>2CO2+2H2O
y-------------3y------2y
=>Ta có :
\(\left\{{}\begin{matrix}x+y=0,04\\2x+3y=0,11\end{matrix}\right.\)
=>x=0,01, y=0,03 mol
=>%VCH4=\(\dfrac{0,01.22,4}{0,896}100\)=25%
=>%VC2H4=75%
=>m CO2=(0,01+0,03.2).44=3,08g
a, ta có PTHH :
CH4 + 2O2 \(\Rightarrow\) CO2 + 2H2O (*)
C2H4 + 3O2 \(\Rightarrow\) 2CO2 + 2H2O(+)
b, Gọi x, y lần lượ là số mol của CH4 , C2H4 ( x, y > 0 )
Ta có : nh = \(\dfrac{0,896}{22,4}=0,04\left(mol\right)\)
\(\Rightarrow x+y=0,04\left(mol\right)\) (1)
lại có no2 pứ = \(\dfrac{3,52}{32}=0,11\left(mol\right)\)
\(\Rightarrow2x+3y=0,11\left(mol\right)\) (2)
Từ (1) và (2) ta có hệ pt:\(\left\{{}\begin{matrix}x+y=0,04\\2x+3y=0,11\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,01\left(mol\right)\\y=0,03\left(mol\right)\end{matrix}\right.\)
%VCH4 = \(\dfrac{0,01}{0,04}\times100\%=25\%\)
%VC2H4 = \(\dfrac{0,03}{0,04}\times100\%=75\%\)
c, Theo pt nCO2(*) = nCH4 = 0,01 (mol)
nCO2(+) = 2nC2H4 = 0,6 (mol)
\(\Rightarrow m_{co_2}=\left(0,01+0,6\right)\times44=26,84\left(g\right)\)