Question

Hòa tan hoàn toàn 6,5g kẽm bằng 250ml dung dịch HCl
a.Viết PT hóa học
b.Tính thể tích khí hiđro thoát ra (ĐKTC)
c.Tính nồng đọ mol dung dịch HCl phản ứng
d.Nếu dùng 250ml dung dịch HCl trên trung hòa V ml dung dịch Ca(OH)2 2M.Tính V?

Answer 1

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

a. \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,1----0,2----------------------0,1

b. \(V_{H_2}=0,1\cdot22,4=2,24l\)

c. \(CM_{HCl}=\dfrac{0,2}{0,25}=0,8\)

d. \(2HCl+Ca\left(OH\right)_2\rightarrow CaCl_2+2H_2O\)

---0,2----------0,1

\(V_{Ca\left(OH\right)_2}=\dfrac{0,1}{2}=0,05l=50ml\)

Answer 2

a) Zn + 2HCl -> ZnCl2 + H2

......0,1.......0,2..........0.1.........0,1(mol)

b) nZn = 6.5/65=0.1(mol)

VH2 = 0,1.22,4 = 2,24l

c)C_M(HCl)=0.2/0.25=0.8(M)

d) Ca(OH)2 + 2 HCL --> CaCl2 + 2 H2O

..........0.1.................0.2...................................................(mol)

V=0.1/2=0.5(l)

Answer 3

a) Zn + 2HCl → ZnCl₂ + H₂ (1)

b) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

Theo Pt1: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow V_{H_2}=0,1\times22,4=2,24\left(l\right)\)

c) Theo Pt1: \(n_{HCl}=2n_{Zn}=2\times0,1=0,2\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,25}=0,8\left(M\right)\)

d) 2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O (2)

Theo Pt: \(n_{Ca\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=\dfrac{1}{2}\times0,2=0,1\left(mol\right)\)

\(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{0,1}{2}=0,05\left(l\right)=50\left(ml\right)\)

Vậy \(V=50\left(ml\right)\)