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\(n_{HCl}=0,2.2=0,4\left(mol\right)\\ a.Zn+2HCl\rightarrow ZnCl_2+H_2\\ b.0,2........0,4.......0,2.......0,2\left(mol\right)\\ m=m_{Zn}=0,2.65=13\left(g\right)\\ c.m_{ZnCl_2}=136.0,2=27,2\left(g\right)\\ d.V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)
a. \(n_{Zn}=\dfrac{6.5}{65}=0,1\left(mol\right)\)
PTHH : Zn + 2HCl -> ZnCl2 + H2
0,1 0,2 0,1
b. \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c. \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)
\(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,1
\(V_{H_2}=0,1\cdot22,4=2,24l\)
\(m_{HCl}=0,2\cdot36,5=7,3g\)
\(a,n_{Zn}=\dfrac{0,65}{65}=0,01\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,01--->0,02---->0,01---->0,01
\(V_{H_2}=0,01.22,4=0,224\left(l\right)\\ b,m_{ZnCl_2}=0,01.136=1,36\left(g\right)\\ V_{ddHCl}=\dfrac{0,02}{2}=0,01\left(l\right)\)
a)
Zn + 2HCl → ZnCl2 + H2
b) nZn = \(\dfrac{3,5}{65}\)=\(\dfrac{7}{130}\) mol
Theo tỉ lệ phản ứng => nH2 = nZn= \(\dfrac{7}{130}\)mol
<=> V H2 = \(\dfrac{7}{130}\).22,4 = 1,206 lít
c) nZnCl2 = nZn => mZnCl2 = \(\dfrac{7}{130}\).136= 7,32 gam
a)
\(Zn + 2HCl \to ZnCl_2 + H_2\)
b),c)
Theo PTHH :
\(n_{ZnCl_2} = n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\)
Vậy :
\(m_{ZnCl_2} = 0,2.136 = 27,2(gam)\\ V_{H_2} =0,2.22,4 = 4,48(lít)\)
a,\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1--------------->0,1------>0,1
b, => \(\left\{{}\begin{matrix}C_{M\left(ZnCl_2\right)}=\dfrac{0,1}{\dfrac{6}{1000}}=\dfrac{50}{3}M\\V_{H_2}=0,1.22,4=2,24\left(l\right)\end{matrix}\right.\)
c, \(n_{O_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: 2H2 + O2 --to--> 2H2O
LTL: \(\dfrac{0,1}{2}< 0,1\)=> O2 dư
Theo pt: \(n_{O_2}=\dfrac{1}{2}n_{H_2}=\dfrac{1}{2}.0,1=0,05\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(dư\right)}=\left(0,1-0,05\right).32=1,6\left(g\right)\\V_{O_2\left(dư\right)}=\left(0,1-0,05\right).22,4=1,12\left(l\right)\end{matrix}\right.\)
nMg = 3,6 : 24 = 0,15 (mol)
pthh : Mg + 2HCl --> MgCl2 + H2
0,15-----------> 0,15 --->0,15 (mol)
mMgCl2 = 0,15 . 95 = 14,25 (mol)
VH2 (đkc)= 0,15. 24,79 = 3,718(l)
\(n_{Mg}=\dfrac{3,6}{24}=0,15mol\)
\(Mg+2HCl\rightarrow MgCl_2+H_2\)
0,15 0,3 0,15 0,15
\(m_{MgCl_2}=0,15\cdot95=14,25g\)
\(V_{H_2}=0,15\cdot22,4=3,36l\)
a. Zn + 2HCl → ZnCl2 + H2
b. nZn = n\(_{ZnCl_2}\) =\(\dfrac{13}{65}=0,2\left(mol\right)\) => m\(_{ZnCl_2}\)= 0,2.136 = 27,2(g)
c. n\(_{H_2}\)= nZn = 0,2 (mol) => V\(_{H_2}\)=0,2.22,4 = 4,48 (lít)