Question

Cho a gam Zn vào 200g đ HCl thoát ra 3,36l khí H2(dktc) A) tìm m gam Zn B)Tìm m zncl2 C) Tính C% đ hcl

Answer 1

\(n_{H2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

     0,15<---0,3<-----0,15<-----0,15

a) \(m_{Zn}=0,15.65=9,75\left(g\right)\)

b) \(m_{ZnCl2}=0,15.136=20,4\)

c) \(C\%_{ddHCl}=\dfrac{0,3.36,5}{200}.100\%=5,475\%\)

Answer 2

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

a, \(n_{Zn}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{Zn}=0,15.65=9,75\left(g\right)\)

b, \(n_{ZnCl_2}=n_{H_2}=0,15\left(mol\right)\Rightarrow m_{ZnCl_2}=0,15.136=20,4\left(g\right)\)

c, \(n_{HCl}=2n_{H_2}=0,3\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,3.36,5}{200}.100\%=5,475\%\)