Question

Hòa tan 9.75 g zn bằng 30 ml dd hcl 20% có d = 1.825 g/ mol . Hãy tính C% của muối zn trong dd

Answer 1

\(Zn+2HCl--->ZnCl_2+H_2\) \(nZn=\dfrac{9,75}{65}=0,15(mol)\) \(mddHCl=DHCl.VddHCl=1,825.30=54,75(g)\) \(\Rightarrow mHCl=\dfrac{20.54,75}{100}=10,95\left(g\right)\) \(\Rightarrow nHCl=\dfrac{10,95}{36,5}=0,3\left(mol\right)\) So sánh: \(\dfrac{nZn}{1}=0,15=\dfrac{nHCl}{2}=0,15\) \(\Rightarrow\)Phản ứng vừa đủ. Dung dich sau phản ứng là: \(ZnCl_2\) Theo PTHH: \(nZnCl_2=0,15(mol) \) \(\Rightarrow mZnCl_2=20,4\left(g\right)\) \(nH_2=0,15\left(mol\right)\) \(\Rightarrow mH_2=0,3\left(g\right)\) \(mddsau=mZn+mddHCl-mH_2\) \(\Leftrightarrow mddHCl=9,75+54,75-0,3=64,2\left(g\right)\) \(C\%ZnCl_2=\dfrac{20,4}{64,2}.100=31,78\%\)

Answer 2

mdd_{HCl 20%}=30 . 1,825=54,75(g)

m_HCl=20 . 54,75/100=10,95(g)

=>C%=9,75/9,75+10,95 . 100=47,1%