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a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
____0,1_____0,2______0,1_____0,1 (mol)
\(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
b, \(C_{M_{HCl}}=\dfrac{0,2}{0,1}=1\left(M\right)\)
c, \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Xét tỉ lệ: \(\dfrac{0,15}{1}>\dfrac{0,1}{1}\), ta được CuO dư.
Theo PT: \(n_{CuO\left(pư\right)}=n_{Cu}=n_{H_2}=0,1\left(mol\right)\Rightarrow n_{CuO\left(dư\right)}=0,05\left(mol\right)\)
⇒ m chất rắn = mCuO (dư) + mCu = 0,05.80 + 0,1.64 = 10,4 (g)
nHCl=0,1.5=0,5(mol)
PTHH: Zn + 2HCl -> ZnCl2 + H2
nZn=nZnCl2=nH2=nHCl/2=0,5/2=0,25(mol)
mZnCl2=0,25.136=34(g)
Mà em ơi không cho khối lượng riêng dd HCl à e,?
\(a,Zn+2HCl\rightarrow ZnCl_2+H_2\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right);n_{HCl}=2.0,2=0,4\left(mol\right)\\ m_{ZnCl_2}=136.0,2=27,2\left(g\right);C_{MddHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\\ b,Zn+CuSO_4\rightarrow ZnSO_4+Cu\\ n_{CuSO_4}=\dfrac{20.10\%}{160}=0,0125\left(mol\right);n_{Zn}=0,2\left(mol\right)\\ Vì:\dfrac{0,0125}{1}< \dfrac{0,2}{1}\Rightarrow Zn.dư\\ n_{Zn\left(p.ứ\right)}=n_{ZnSO_4}=n_{CuSO_4}=0,0125\left(mol\right)\\m_{Zn\left(p.ứ\right)}=0,0125.65=0,8125\left(g\right)\\ m_{ddsau}=m_{Zn\left(p.ứ\right)}+m_{ddCuSO_4}=0,8125+20=20,8125\left(g\right)\\ C\%_{ddZnSO_4}=\dfrac{0,0125.161}{20,8125}.100\approx9,67\%\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right);n_{HCl}=0,5.1=0,5\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,5}{2}>\dfrac{0,1}{1}\Rightarrow Zn.hết,HCldư\\ n_{HCl\left(dư\right)}=0,5-2.0,1=0,3\left(mol\right)\\ m_{HCl\left(dư\right)}=0,3.36,5=10,95\left(g\right)\)
\(a.PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ b.n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ PTHH:Zn+2HCl\xrightarrow[]{}ZnCl_2+H_2\\ n_{H_2}=0,2.2=0,4\left(mol\right)\\ V_{H_2}=0,4.22,4=8,96\left(l\right)\\ c.n_{HCl}=n_{Zn}=0,2mol\\ C_{MHCl}=\dfrac{0,4}{0,1}=4\left(M\right)\)
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
PTHH :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,2 0,4 0,2 0,2
\(b,V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(c,C_M=\dfrac{n}{V}=\dfrac{0,4}{0,1}=4M\)
`a)PTPƯ: Zn + 2HCl -> ZnCl_2 + H_2↑`
____________________________________________
`b) n_[Zn] = 13 / 65 = 0,2 (mol)`
Theo `PTPƯ` có: `n_[HCl] = 2n_[Zn] = 2 . 0,2 = 0,4 (mol)`
`-> m_[dd HCl] = [ 0,4 . 36,5 ] / [ 7,3 ] . 100 = 200 (g)`
_____________________________________________
`c)` Theo `PTPƯ` có: `n_[H_2] = n_[ZnCl_2] = n_[Zn] = 0,2 (mol)`
`-> C%_[ZnCl_2] = [ 0,2 . 136 ] / [ 13 + 200 - 0,2 . 2 ] . 100 ~~ 12,79 %`
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
0,1 0,1 0,1
\(b,C_M=\dfrac{0,1}{0,1}=1M\)
\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)
nHCl=0,1. 0,5=0,05(mol)
PTHH: Zn + 2 HCl -> ZnCl2 + H2
0,025<-----0,05--------0,025-----0,025(mol)
mZnCl2=0,025 x 136 = 3,4(g)
mddZnCl2= mZn + mddHCl - mH2 (Nhma chưa cho khối lượng riêng của ddHCl nè :) )