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a)
$n_{HCl} = \dfrac{150.14,6\%}{36,5} = 0,6(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH :
$n_{H_2} = 0,3(mol)$
$n_{Al} = n_{AlCl_3} = \dfrac{1}{3}n_{HCl} = 0,2(mol)$
$m_{Al} = 0,2.27 = 5,4(gam)$
b)
$m_{dd} = 5,4 + 150 - 0,3.2 = 154,8(gam)$
$C\%_{AlCl_3} = \dfrac{0,2.133,5}{154,8}.100\% = 17,25\%$
\(n_{H_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.1..........0.3.......0.1...........0.15\)
\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)
\(C\%_{HCl}=\dfrac{10.95}{150}\cdot100\%=7.3\%\)
\(m_{Al}=0.1\cdot27=2.7\left(g\right)\)
Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
_____0,1____0,3____________0,15 (mol)
b, mHCl = 0,3.36,5 = 10,95 (g)
c, \(C\%_{HCl}=\dfrac{10,95}{150}.100\%=7,3\%\)
d, mAl = 0,1.27 = 2,7 (g)
Bạn tham khảo nhé!
\(a,n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
PTHH: \(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
0,05<-0,1<------0,05<---0,05
\(b,m_{Fe}=0,05.56=2,8\left(g\right)\\ c,m_{ddHCl}=\dfrac{0,1.36,5}{14,6\%}=25\left(g\right)\\ m_{dd}=25+2,8-0,05.2=27,7\left(g\right)\\ \rightarrow C\%_{FeCl_2}=\dfrac{0,05.127}{27,7}.100\%=22,92\%\)
Theo gt ta có: $n_{Al}=0,1(mol)$
a, $2Al+6HCl\rightarrow 2AlCl_3+3H_2$
b, $\Rightarrow n_{H_2}=0,15(mol)\Rightarrow V_{H_2}=3,36(l)$
c, Ta có: $n_{HCl}=0,3(mol)\Rightarrow m_{HCl}=10,95(g)\Rightarrow \%m_{ddHCl}=219(g)$
d, Bảo toàn khối lượng ta có: $m_{dd}=221,4(g)$
$\Rightarrow \%C_{AlCl_3}=6,02\%$
a: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
b: \(n_{H2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)
\(\Leftrightarrow n_{Al}=0.1\left(mol\right)\)
\(m_{Al}=n_{Al}\cdot M_{Al}=0.1\cdot27=2.7\left(g\right)\)
Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)
PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
____0,5____________0,5 (mol)
a, \(m_{CuCl_2}=0,5.135=67,5\left(g\right)\)
b, Có: m dd sau pư = mCuO + m dd HCl = 40 + 200 = 240 (g)
\(\Rightarrow C\%_{CuCl_2}=\dfrac{67,5}{240}.100\%=28,125\%\)
Bạn tham khảo nhé!
\(m_{HCl}=150\cdot14.6=21.9\left(g\right)\)
\(n_{HCl}=\dfrac{21.9}{36.5}=0.6\left(mol\right)\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(0.2.........0.6..........0.2.......0.3\)
\(m_{Al}=0.2\cdot27=5.4\left(g\right)\)
\(m_{dd}=5.4+150-0.3\cdot2=154.8\left(g\right)\)
\(C\%_{AlCl_3}=\dfrac{0.2\cdot133.5}{154.8}\cdot100\%=17.24\%\)
2Al+ 6HCl --------> 2AlCl3 + 3H2
\(n_{HCl}=\dfrac{150.14,6\%}{36,5}=0,6\left(mol\right)\)
Ta có : \(n_{Al}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\)
=> \(m_{Al}=0,2.27=5,4\left(g\right)\)
\(m_{ddsaupu}=5,4+150-0,6.2=154,2\left(g\right)\)
=>\(C\%_{AlCl_3}=\dfrac{0,2.133,5}{154,2}.100=17,32\%\)