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nHCl=0,2(mol)
nH2= 1,12/22,4=0,05(mol)
a) PTHH: 2Al + 6 HCl -> 2AlCl3 + 3 H2
b) Ta có: 0,2/6 > 0,05/3
=> H2 hết, HCl dư, tính theo nH2
nHCl(dư)= 0,2 - 6/3 . 0,05=0,1(mol)
=> mHCl(dư)=36,5.0,1=3,65(g)
c) nAlCl3= 2/3. 0,05=1/30(mol)
=> mAlCl3= 1/30. 133,5=4,45(g)
a)
\(n_{H_2} = \dfrac{1,12}{22,4} = 0,05(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\)
Ta thấy :
\(n_{HCl} = 0,2 > 2n_{H_2} = 0,1\) nên HCl dư.
Theo PTHH : \(n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{1}{30}(mol)\\ \Rightarrow m_{Al} = \dfrac{1}{30}.27 = 0,9(gam)\)
Ta có :
\(n_{HCl\ pư} = 2n_{H_2} = 0,1(mol)\\ \Rightarrow n_{HCl\ dư} = 0,2 - 0,1 = 0,1(mol)\\ \Rightarrow m_{HCl\ dư} = 0,1.36,5 = 3,65(gam)\)
Ta có: \(n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
\(a,PTHH:2Al+6HCl--->2AlCl_3+3H_2\uparrow\)
Theo PT: \(n_{Al}=\dfrac{2}{3}.n_{H_2}=\dfrac{2}{3}.0,05=\dfrac{1}{30}\left(mol\right)\)
\(\Rightarrow m_{Al}=\dfrac{1}{30}.27=0,9\left(g\right)\)
b. Ta thấy: \(\dfrac{\dfrac{1}{30}}{2}>\dfrac{0,02}{6}\)
Vậy nhôm dư.
2Al+6HCl->2AlCl3+3H2
0,05----0,15------------------0,075 mol
n H2=\(\dfrac{1,68}{22,4}\)=0,075 mol
=>m Al=0,05.27=1,35g
=>HCl dư =>m HCl=0,1.36,5=3,65g
2Al + 6HCl -> 2AlCl3 + 3H2
nH2=0,05(mol)
Vì 0,05.2<0,2 nên sau PƯ HCl dư 0,1 mol
Theo PTHH ta có:
nAl=\(\dfrac{2}{3}\)nH2=\(\dfrac{0,1}{3}\left(mol\right)\)
mAl=\(\dfrac{0,1}{3}\).27=0,9(g)
mHCl dư=36,5.0,1=3,65(g)
a, Ta có pt pư
\(Fe+H_2SO_4-->FeSO_4+H_2\)
Ta có
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
=> \(H_2SO_4\) dư
\(m_{H_2SO_4}=0,15\cdot98=14,7\left(g\right)\)
\(\Rightarrow m_{dư\left(H_2SO_4\right)}=19,6-14,7=4,9\left(g\right)\)
b,
Ta có
\(m_{Fe}=0,15\cdot56=8,4\left(g\right)\)
\(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
a) 2Al + 6HCl --> 2AlCl3 + 3H2
b) \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
2Al + 6HCl --> 2AlCl3 + 3H2
\(\dfrac{0,4}{3}\)<--------------------0,2
=> Al dư
\(m_{Al\left(dư\right)}=\left(0,2-\dfrac{0,4}{3}\right).27=1,8\left(g\right)\)
\(V_{H_2}\)= \(\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3mol\)PTHH: 2Al+6HCL→2AlCl3+3H2a)Theo pt: \(n_{Al_{ }}=\dfrac{2}{3}n_{H_2}=\dfrac{2}{3}.0,3=0,2mol\)⇒ mAl=n.M=0,2.27=5,4gb)Theo pt:\(n_{HCl}=\dfrac{6}{2}n_{H_2}=\dfrac{6}{2}.0,3=0,9mol\)⇒mHCl=n.M=0,9.36.5=32,85g
\(n_{HCl\left(bđ\right)}=2\left(mol\right);n_{H_2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\\ 2Al+6HCl\rightarrow2AlCl_3+3H_2\\ Vì:\dfrac{2}{6}>\dfrac{0,05}{3}\Rightarrow HCldư\\ a,n_{Al}=\dfrac{2}{3}.0,05=\dfrac{1}{30}\left(mol\right)\\ m_{Al}=\dfrac{1}{30}.27=0,9\left(g\right)\\ b,n_{HCl\left(dư\right)}=2-0,05.2=1,9\left(mol\right)\)