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a,\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl → ZnCl2 + H2
Mol: 0,1 0,1 0,1
b,\(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
c,\(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
a)PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(n_{ZnCl_2}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=n.M=0,1.136=13,6\left(g\right)\)
c) \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\)
\(V_{H_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)
a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Zn + 2HCl -----> ZnCl2 + H2
0,2 0,4 0,2 0,2
b, \(m_{Zn}=0,2.65=13\left(g\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
c, \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
\(a,n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,2<--0,4<--------0,2<---0,2
\(b,\left\{{}\begin{matrix}m_{Zn}=0,1.65=13\left(g\right)\\m_{HCl}=0,4.36,5=14,6\left(g\right)\end{matrix}\right.\\ c,m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
a. \(n_{Zn}=\dfrac{19.5}{65}=0,3\left(mol\right)\)
\(n_{HCl}=\dfrac{14.6}{36.5}=0,4\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,4 0,2 0,2
Ta thấy : \(\dfrac{0.3}{1}>\dfrac{0.4}{2}\) => Zn dư , HCl đủ
b. \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c. \(m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
a) Zn + 2HCl --> ZnCl2 + H2 ↑ (1)
0,3 -->0,15 -->0,15 (mol)
nZn= 19,5/65 = 0,3 mol
nHCl= 14,5/37,5 = 0,3 mol
Ta có : nZn bài ra / nZn phương trình=0,3/1=0,3 (mol)
nHCl bài ra / nHCl phương trình=0,3/2=0,15 (mol)
=> HCl đủ,Zn dư
b) Theo PT(1) => nH2=0,15(mol)
=>VH2=0,15 x 22,4 = 3,36(l)
c) Theo PT(1) => nZnCl2=0,15(mol)
=>mZnCl2=0,15 x 136 = 20,4(g)
\(n_{Zn}=\dfrac{3,25}{65}=0,05mol\)
a)\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,05 0,1 0,05 0,05
b)\(m_{ZnCl_2}=0,05\cdot136=6,8g\)
c)\(V_{H_2}=0,05\cdot22,4=1,12l\)
\(m_{HCl}=219.10\%=21,9\left(g\right)\Rightarrow n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
Theo PT: \(n_{ZnCl_2}=n_{H_2}=\dfrac{1}{2}n_{HCl}=0,3\left(mol\right)\)
a, \(V_{H_2}=0,3.24,79=7,437\left(l\right)\)
b, \(m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
\(a,Zn+2HCl\to ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2(mol);n_{HCl}=\dfrac{21,9}{36,5}=0,6(mol)\)
Vì \(\dfrac{n_{Zn}}{1}<\dfrac{n_{HCl}}{2}\) nên \(HCl\) dư
\(n_{HCl(dư)}=0,6-0,2.2=0,2(mol)\\ c,n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,2(mol)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2(g)\\ V_{H_2}=0,2.22,4=4,48(l)\)
Zn +2 HCl ---> ZnCl2 + H2
0,1-----0,2----------0,1-------------0,1 mol
ZnO + 2HCl ---> ZnCl2 + H2O
0,2------0,4-------0,2--------0,2
n H2=\(\dfrac{2,24}{22,4}=0,1mol\)
=>m Zn=0,1.65=6,5g
=>m HCl(1)=0,2.36,5=7,3g
=>m HCl(2)=14,6g -> nHCl=0,4 mol
=>%m Zn=\(\dfrac{6,5}{6,5+14,4}.100=31,1\%\)
=>%m ZnO=68,9%
b)
->m HCl=0,6.36,5=21,9g
->m ZnCl2=0,3.136=40,8g
a) Zn+2HCl ---------> ZnCl2 +H2
b) n H2=5,6/22,4=0,25(mol)
Zn+2HCl ---------->ZnCl2 +H2
TPT:1. 2. 1. 1
TB: ? ? ? 0,25 Theo phương trình và bài ra
ta có:
n HCl= 0,25×2/1=0,5(mol)
m HCl=36,5×0,5=18,25(g)
c) ta có :
n ZnCl2 = 0,25×1/1=0,25(mol)
m ZnCl2=136×0,25=34(mol)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{ZnCl_2}=0,25mol\\n_{HCl}=0,5mol\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl}=0,5\cdot36,5=18,25\left(g\right)\\m_{ZnCl_2}=0,25\cdot136=34\left(g\right)\end{matrix}\right.\)