Question

Answer 1

a) $Fe + 2HCl \to FeCl_2 + H_2$
Theo PTHH : 

$n_{FeCl_2} = n_{H_2} = n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$V_{H_2} = 0,2.22,4 = 4,48(lít)$

b) $m_{FeCl_2} = 0,2.127 =25,4(gam)$

c) $n_{HCl} = 2n_{Fe} = 0,2.2 = 0,4(mol) \Rightarrow C_{M_{HCl}} = \dfrac{0,4}{0,1} = 4M$

d) $CuO + H_2 \xrightarrow{t^o} Cu + H_2O$
Ta thấy : $n_{CuO} = \dfrac{48}{80}  =0,6 > n_{H_2} = 0,2$ nên CuO dư

$n_{Cu} = n_{H_2} = 0,2(mol)$
$m_{Cu} = 0,2.64 = 12,8(gam)$

Answer 2

n_Fe = \(\dfrac{11,2}{56}=0,2\left(mol\right)\)

PTHH: Fe + 2HCl ---> FeCl₂ + H₂

a. Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\)

=> \(V_{H_2}=0,2.22,4=4,48\left(lít\right)\)

b. Theo PT: \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\)

=> \(m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

c. Theo PT: n_HCl = 2.n_Fe = 2.0,2 = 0,4(mol)

Đổi 100ml = 0,1(lít)

=> C_M = \(\dfrac{0,4}{0,1}=4\)(g/mol)

d. PT: H₂ + CuO ---t^o---> Cu + H₂O

Ta có: n_CuO = \(\dfrac{48}{80}=0,6\left(mol\right)\)

Theo PT: n_Cu = n_CuO = 0,6(mol)

=> m_Cu = 0,6 . 64 = 38,4(g)