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\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right);n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\a, Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,V\text{ì}:\dfrac{0,2}{1}>\dfrac{0,1}{1}\Rightarrow Zn\text{dư}\\ \Rightarrow n_{Zn\left(p.\text{ứ}\right)}=n_{ZnCl_2}=n_{H_2}=0,1\left(mol\right)\\b, m_{Zn\left(p.\text{ứ}\right)}=0,1.65=6,5\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ m_{HCl}=0,2.36,5=7,3\left(g\right)\\ d,m_{ZnCl_2}=136.0,1=13,6\left(g\right)\)
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
b) Ta có: \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,8mol\\n_{H_2}=0,4mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,8\cdot36,5=29,2\left(g\right)\\V_{H_2}=0,4\cdot22,4=8,96\left(l\right)\end{matrix}\right.\)
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(b,n_{Zn}=\dfrac{m}{M}=\dfrac{16,25}{65}=0,25\left(mol\right)\\ Theo.PTHH:n_{HCl}=2.n_{Zn}=2.0,25=0,5\left(mol\right)\\ m_{HCl}=n.M=0,5.36,5=18,25\left(g\right)\)
\(Theo.PTHH:n_{H_2}=n_{Zn}=0,25\left(mol\right)\\ V_{H_2\left(đktc\right)}=n.22,4=0,25.22,4=5,6\left(l\right)\)
a)PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b)Khối lượng Zn:\(m_{Zn}=\dfrac{16,25}{65}=0,25\left(mol\right)\)
Ta có: \(n_{HCl}=2n_{Zn}=0,5\left(mol\right)\)
Khối lượng axit HCl cần dùng là: \(m_{HCl}=0,5.36,5=18,25\left(g\right)\)
c)Theo pt ta có: \(n_{H_2}=n_{Zn}=0,25\left(mol\right)\)
Thể tích H2 là: \(V_{H_2}=n.22,4=0,25.22,4=5,6\left(ml\right)\)
a) Zn + 2HCl →ZnCl2 + H2
b) nZn = 6,5/65 = 0,1 mol . Theo tỉ lệ pư => nH2 = nZn = nZnCl2 =0,1 mol <=> VH2(đktc) = 0,1.22,4 = 2,24 lít.
c) mZnCl2 = 0,1 . 136 = 13,6 gam
d) nHCl =2nZn = 0,2 mol => mHCl = 0,2.36,5= 7,3 gam
Cách 2: áp dụng định luật BTKL => mHCl = mZnCl2 + mH2 - mZn
<=> mHCl = 13,6 + 0,1.2 - 6,5 = 7,3 gam
Theo gt ta có: $n_{Zn}=0,1(mol)$
a, $Zn+2HCl\rightarrow ZnCl_2+H_2$
b, Ta có: $n_{H_2}=0,1(mol)\Rightarrow V_{H_2}=2.24(l)$
c, Ta có: $n_{HCl}=2.n_{Zn}=0,2(mol)\Rightarrow m_{HCl}=7,3(g)$
a+b+c) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{ZnCl_2}=n_{H_2}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\V_{H_2}=0,1\cdot22,4=2,24\left(l\right)\end{matrix}\right.\)
d) PTHH: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
Theo PTHH: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)
a) PTHH: Zn+2HCl-------to----> ZnCl2+H2
nZn=\(\dfrac{m}{M}=\dfrac{32,5}{65}=0,5\left(mol\right)\)
---> \(n_{H_2}\)=0,5(mol)
b)\(V_{H_2}\)=n.22,4=0,5.22,4=11,2(lít)
c)\(n_{ZnCl_2}\)=0,5(mol)
\(m_{ZnCl_2}\)=n.M=0,5.136=68(gam)
nZn= 13/65=0,2(mol)
a) PTHH: Zn + 2 HCl -> ZnCl2 + H2
b) nH2=nZnCl2=nZn=0,2(mol)
=>V(H2,đktc)=0,2 x 22,4= 4,48(l)
c) khối lượng muối sau phản ứng chứ nhỉ?
mZnCl2=136.0,2=27,2(g)
a) Zn + 2HCl --> ZnCl2 + H2
b) \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
PTHH: Zn + 2HCl --> ZnCl2 + H2
_____0,3<---0,6<----------------0,3
=> mZn = 65.0,3 = 19,5(g)
c) mHCl = 0,6.36,5 = 21,9(g)