Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
PTHH: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\uparrow\)
Ta có: \(n_{H_2SO_4}=0,3\cdot1=0,3\left(mol\right)=n_{Zn}=n_{ZnSO_4}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Zn}=0,3\cdot65=19,5\left(g\right)\\V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\C_{M_{ZnSO_4}}=\dfrac{0,3}{0,3}=1\left(M\right)\end{matrix}\right.\)
a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1--->0,2------->0,1----->0,1
VH2 = 0,1.22,4 = 2,24 (l)
b, \(C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\)
c, \(C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{0,2}=0,5M\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,4 0,8 0,4 0,4
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\\ b,C\%_{HCl}=\dfrac{0,8.36,5}{150}.100\%=19,5\%\\ c,m_{\text{dd}}=26+150-\left(0,4.2\right)=175,2\left(g\right)\\ C\%_{ZnCl_2}=\dfrac{0,4.136}{175,2}.100\%=31\%\)
`n_[Zn]=13/65=0,2(mol)`
`Zn + 2HCl -> ZnCl_2 + H_2 \uparrow`
`0,2` `0,4` `0,2` `(mol)`
`a)V_[H_2]=0,2.22,4=4,48(l)`
`b)C%_[HCl]=[0,4.36,5]/150 .100~~9,73%`
\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{ZnCl_2}=n_{Zn}=n_{H_2}=0,15\left(mol\right);n_{HCl}=2.0,15=0,3\left(mol\right)\\ a,m=m_{Zn}=0,15.65=9,75\left(g\right)\\ b,C_{MddHCl}=\dfrac{0,3}{0,15}=0,2\left(l\right)\\ c,m_{ZnCl_2}=0,15.136=20,4\left(g\right)\)
a, nH2=5,6/22,4=0,25 mol
Zn+2HCl->ZnCl2+H2
0,25 0,5 0,25 0,25
mZn pư=0,25.65=16,25 g
b, C%HCl=0,5.36,5.100/200=9,125%
Zn + 2HCl => ZnCl2 + H2
nHCl = 0.2x2 = 0.4 (mol)
Theo pt => nZn = 0.2 (mol) = nH2
mZn = n.M = 0.2 x 65 = 13 (g)
VH2 = 22.4 x n = 22.4 x 0.2 = 4.48 (l)
nZnCl2 = 0.2 (mol); 200 ml = 0.2 (l)
CM dd sau pứ = 0.2/0.2 = 1M