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a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
____0,15<--0,3<-------------0,15
=> mFe = 0,15.56 = 8,4 (g)
b) \(C_{M\left(ddHCl\right)}=\dfrac{0,3}{0,05}=6M\)
\(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)
PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
a, \(n_{H_2}=n_{Mg}=0,5\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,5.22,4=11,2\left(l\right)\)
b, \(n_{H_2SO_4}=n_{Mg}=0,5\left(mol\right)\)\(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,5}{0,2}=2,5\left(M\right)\)
a)
$Fe + 2HCl \to FeCl_2 + H_2$
b) Theo PTHH : $n_{Fe} = n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$m_{Fe} = 0,15.56 = 8,4(gam)$
c) $n_{HCl} = 2n_{H_2} = 0,3(mol)$
$\Rightarrow C_{M_{HCl}} = \dfrac{0,3}{0,05} = 6M$
d) $2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{H_2SO_4} = \dfrac{1}{2}n_{NaOH} = 0,25(mol)$
$m_{dd\ H_2SO_4} = \dfrac{0,25.98}{20\%} = 122,5(gam)$
$V_{dd\ H_2SO_4} = \dfrac{122,5}{1,14} = 107,5(ml)$
a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,15\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,15.56=8,4\left(g\right)\)
a) `n_{H_2} = (3,36)/(22,4) = 0,15 (mol)`
PTHH: `Fe + 2HCl -> FeCl_2 + H_2`
Theo PT: `n_{Fe} = n_{H_2} = 0,15 (mol)`
`=> m_{Fe} = 0,15.56 = 8,4 (g)`
b) Theo PT: `n_{HCl} = 2n_{H_2} = 0,3 (mol)`
`=> m_{ddHCl} = (0,3.36,5)/(16\%) = 68,4375 (g)`
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
a)PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
b) \(n_{Mg}=n_{H_2}=0,15\left(mol\right)\)
\(m_{Mg}=0,15.24=3,6\left(g\right)\)
c) \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\)
\(C_{M_{HCl}}=\dfrac{0,3}{0,5}=0,6\left(M\right)\)
\(n_{HCl}=0,3.2=0,6\left(mol\right)\)
PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
a, \(n_{Al}=\dfrac{1}{3}n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow m_{Al}=0,2.27=5,4\left(g\right)\)
b, \(n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)
\(\Rightarrow V_{H_2}=0,3.22,4=6,72\left(l\right)\)