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Ta có: \(n_{H_2}=\dfrac{74,37}{24,79}=3\left(mol\right)\)
PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)
a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=2\left(mol\right)\)
\(\Rightarrow m_{Al}=2.27=54\left(g\right)\)
b, \(n_{H_2SO_4}=n_{H_2}=3\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=3.98=294\left(g\right)\)
a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,6\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)
b, Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
\(PTHH:2Al+6HCl->2AlCl_3+3H_2\)
0,2<--0,6<----------0,2<------0,3 (mol)
\(n_{H_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
\(m_{HCl}=n\cdot M=0,6\cdot\left(1+35,5\right)=21,9\left(g\right)\)
\(m_{AlCl_3}=n\cdot M=0,2\cdot\left(27+35,5\cdot3\right)=26,7\left(g\right)\)
a, PT: 2Al+6HCl→2AlCl3+3H2
Ta có: nH2=6,7222,4=0,3(mol)
Theo PT: nHCl=2nH2=0,6(mol)
⇒mHCl=0,6.36,5=21,9(g)
b, Theo PT: nAl=23nH2=0,2(mol)
⇒mAl=0,2.27=5,4(g)
\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
0,6 1,2 0,6 (mol)
a, mFe = 0,6.56 = 33,6 (g)
b, mHCl = 1,2.36,5 = 43,8 (g)
a, \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
b, \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)
c, \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
PTHH: 2Al + 6HCl --> 2AlCl3 + 3H2
_____0,1--->0,3
=> mHCl = 0,3.36,5 = 10,95(g)
$PTHH:Zn+2HCl\to ZnCl_2+H_2\uparrow$
$n_{Zn}=\dfrac{13}{65}=0,2(mol)$
Theo PT: $n_{ZnCl_2}=n_{H_2}=0,2(mol);n_{HCl}=0,4(mol)$
$a)m_{axit}=m_{HCl}=n.M=0,4.36,5=14,6(g)$
$b)m_{ZnCl_2}=n.M=0,2.136=27,2(g)$
$c)V_{H_2(đktc)}=n.22,4=0,2.22,4=4,48(lít)$
Số mol kẽm là :
\(n=\dfrac{m}{M}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH : Zn + 2HCL -> ZnCl2 + H2
1 2 1 1
0,2 mol -> 0,4 mol 0,2 mol 0,2 mol
a, Khối lượng HCL là :
\(m=n.M=0,4.35,5=14,2\left(g\right)\)
b, Khối lượng ZnCL2 là :
\(m=n.M=0,1.136=13,6\left(g\right)\)
c, Thể tích H2 là : V = n . 22,4 = \(0,1.22,4=2,24\left(l\right)\)
\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=0,4\left(mol\right)\Rightarrow m_{Al}=0,4.27=10,8\left(g\right)\)
b, \(n_{HCl}=2n_{H_2}=1,2\left(mol\right)\Rightarrow m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(n_{H_2}=\dfrac{14,874}{24,79}=0,6\left(mol\right)\)
PTHH :
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,4 1,2 0, 4 0,6
\(m_{Al}=0,4.27=10,8\left(g\right)\)
\(m_{HCl}=1,2.36,5=43,8\left(g\right)\)