Question

Answer 1

Câu 12 :

\(C\%_{NaCl} = \dfrac{8}{40}.100\% = 20\%\)

Câu 13 :

\(a) Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = n_{Zn} = \dfrac{6,5}{65} = 0,1(mol)\\ V_{H_2} = 0,1.22,4 = 2,24(lít)\\ b) n_{HCl} = 2n_{Zn} = 0,2(mol)\\ \Rightarrow C_{M_{HCl}} = \dfrac{0,2}{0,1} = 2M\)

Answer 2

Câu 12 : 

C%NaCl = 8/40 * 100% = 20%

Câu 13: 

nZn = 6.5/65 = 0.1 (mol) 

Zn + 2HCl => ZnCl2 + H2 

0.1.....0.2......................0.1

VddHCl = 0.2 / 0.1 = 2 (l) 

VH2 = 0.1*22.4 = 2.24 (l)