Question

Answer 1

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

a. PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

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

=> m_Zn = 0,2 . 65 = 13(g)

c. Theo PT: \(n_{HCl}=2.n_{Zn}=2.0,2=0,4\left(mol\right)\)

Đổi 100ml = 0,1 lít

=> \(C_{M_{HCl}}=\dfrac{0,4}{0,1}=4M\)

Answer 2

a)Zn + 2HCl --> Zncl2 + H2

Theo Pư: nZn = nZnCl2 = nH2 = \(\dfrac{4.48}{22.4}\) = 0.2(mol)

                   nHCl = 2nH2 = 0.4(mol)

b) mZn = 0.2*64 = 12.8(g)

c)Cm(dd HCl) = \(\dfrac{0.4}{100\cdot10^{-3}}\) = 4(M)