Question

cho mik xin đáp án vs ạ

 

Answer 1

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)

\(n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,25}{1}< \dfrac{0,6}{2}\), ta được HCl dư.

Theo PT: \(n_{ZnCl_2}=n_{Zn}=0,25\left(mol\right)\)

\(\Rightarrow m_{ZnCl_2}=0,25.136=34\left(g\right)\)

c, Theo PT: \(n_{HCl\left(pư\right)}=2n_{Zn}=0,5\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,6-0,5=0,1\left(mol\right)\)

\(\Rightarrow m_{HCl\left(dư\right)}=0,1.36,5=3,65\left(g\right)\)

m_ZnCl2 = 34 (g) (theo phần b)