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Sửa đề: Cho \(65g\) kẽm
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{H_2}=\dfrac{24,79}{24,79}=1(mol)\\ \Rightarrow m_{H_2}=1.2=2(g)\\ \text {Bảo toàn KL}:m_{HCl}=m_{ZnCl_2}+m_{H_2}-m_{Zn}=136+2-65=73(g)\)
\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{H_2}=\dfrac{24,79}{24,79}=1(mol)\\ \Rightarrow m_{H_2}=1.2=2(g)\\ \text {Bảo toàn KL: }m_{HCl}+m_{Zn}=m_{ZnCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}=136+24-2=140,5(g)\\ c,PTHH:H_2+CO_2\xrightarrow{t^o}CO+H_2O\\ H_2+Cl_2\xrightarrow{t^o}2HCl\)
Vì \(\dfrac{n_{H_2}}{1}>\dfrac{n_{CO_2}}{1};\dfrac{n_{H_2}}{1}>\dfrac{n_{Cl_2}}{1}\) nên sau phản ứng \(H_2\) dư
\(\Rightarrow \begin{cases} n_{CO}=0,5(mol\\ n_{HCl}=2n_{Cl_2}=0,7(mol) \end{cases}\\ \Rightarrow m_{hh}=m_{CO}+m{HCl}=0,5.28+0,7.36,5=39,55(g)\\ V_{hh}=V_{CO}+V_{HCl}=0,5.22,4+0,7.22,4=26,88(l)\)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)
PTHH: Zn + 2HCl ---> ZnCl2 + H2
0,1---->0,1----------------->0,1
=> \(\left\{{}\begin{matrix}V_{H_2}=0,1.22,4=2,24\left(l\right)\\C_{M\left(HCl\right)}=\dfrac{0,2}{0,2}=1M\end{matrix}\right.\)
PTHH: CuO + H2 --to--> Cu + H2O
LTL: 0,25 > 0,1 => CuO dư
Theo pthh: nCu = nH2 = 0,1 (mol)
=> mCu = 0,1.64 = 6,4 (g)
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\
pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1 0,2 0,2
\(V_{H_2}=0,2.22,4=4,48l\\
C_M=\dfrac{0,2}{0,2}=1M\\
n_{CuO}=\dfrac{20}{80}=0,25\left(G\right)\\
pthh:CuO+H_2\underrightarrow{t^o}Cu+H_2O\\
LTL:0,25>0,1\)
=>CuO dư
\(n_{Cu}=n_{H_2}=0,1\left(mol\right)\\
m_{Cu}=0,1.64=6,4g\)
\(a,n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1--->0,2------>0,1----->0,1
\(\rightarrow m_{ddHCl}=\dfrac{0,2.36,5}{10\%}=73\left(g\right)\\ b,m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\ V_{H_2}=0,1.22,4=2,24\left(l\right)\)
\(nZn=\dfrac{6,5}{65}=0,1\left(mol\right)\)
\(Zn+2HCl->ZnCl_2+H_2\)
0,1 0,2 0,1 0,1 (mol)
\(mHCl=0,2.36,5=7,3\left(g\right)\)
=> \(mddHCl=\dfrac{7,3.100}{10}=73\left(g\right)\)
mZnCl2 = 0,1 . 136 = 13,6 )g_
VH2 = 0,1 . 22,4 = 2,24 (l)
a) \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
0,1-------------->0,1------>0,1
\(\Rightarrow m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)
b) \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)
c) \(CuO+H_2\xrightarrow[]{t^o}Cu+H_2O\)
0,1<---0,1
\(\Rightarrow m_{CuO}=0,1.80=8\left(g\right)\)
a: Zn+2HCl->ZnCl2+H2
0,2 0,4 0,2 0,2
mZnCl2=0,2*136=27,2(g)
b: V=0,2*22,4=4,48(lít)
Bài 1:
\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ \left(mol\right)....0,1\rightarrow0,2.........0,1.......0,1\\ a,m_{HCl}=0,1.36,5=3,65\left(g\right)\\ b,m_{ZnCl_2}=0,1.136=13,6\left(g\right)\\c,V_{H_2} =0,1.22,4=2,24\left(l\right)\)
Bài 2:
\(n_P=\dfrac{3,1}{31}=0,1\left(mol\right)\\ PTHH:4P+5O_2\underrightarrow{t^o}2P_2O_5\\ \left(mol\right)...0,1\rightarrow0,125...0,05\\ a,m_{P_2O_5}=0,05.142=7,1\left(g\right)\\ a,V_{O_2}=0,125.22,4=2,8\left(l\right)\)
nKClO3 = 49 : 122,5 =0,4(mol)
a) pthh : 2KClO3 -t--> 2KCl + 3O2
0,4------------>0,4----->0,6(mol)
mKCl = 0,4.74,5=29,8 (g)
VO2= 0,6.22,4= 13,44 (l)
câu 2
a nZn = 6,5:65=0,1(mol)
pthh : Zn +2HCl ---> ZnCl2 + H2
0,1->0,2----------------->0,1(mol)
=> VH2 = 0,1.22,4 =2,24(l)
=> mHCl = 0,2 . 36,5=7,3 (g)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right);n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,4}{1}>\dfrac{0,6}{2}\Rightarrow Zn.dư\\ n_{H_2}=n_{Zn\left(p.ứ\right)}=\dfrac{0,6}{2}=0,3\left(mol\right)\\ a,V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ b,n_{Zn\left(dư\right)}=0,4-0,3=0,1\left(mol\right)\Rightarrow m_{Zn\left(dư\right)}=0,1.65=6,5\left(g\right)\)
`n_(Zn)=m/M=(26)/65=0,4(mol)`
`n_(HCl)=m/M=(21,9)/36,5=0,6(mol)`
`PTHH:Zn+2HCl->ZnCl_2 +H_2`
tỉ lệ: 1 ; 2 : 1 : 1
n(mol) 0,3<----0,6---->0,3----->0,3
\(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\left(\dfrac{0,4}{1}>\dfrac{0,6}{2}\right)\)
`=>` `Zn` dư, `HCl` hết, tính theo `HCl`
`V_(H_2)=n*22,4=0,3*22,4=6,72(l)`
`n_(Zn(dư))=0,4-0,3=0,1(mol)`
`m_(Zn(dư))=n*M=0,1*65=6,5(g)`
a.\(n_{HCl}=\dfrac{10,95}{36,5}=0,3mol\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,3 0,15 ( mol )
\(V_{H_2}=0,15.22,4=3,36l\)
b.\(n_{Fe_2O_3}=\dfrac{12}{160}=0,075mol\)
\(Fe_2O_3+3H_2\rightarrow\left(t^o\right)2Fe+3H_2O\)
\(\dfrac{0,075}{1}\) > \(\dfrac{0,15}{3}\) ( mol )
0,15 0,1 ( mol )
\(m_{Fe}=0,1.56=5,6g\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
1 : 6 : 2 : 3 (mol)
0,05 : 0,3 : 0,1 : 0,15 (mol)
\(n_{HCl}=\dfrac{m}{M}=\dfrac{10,95}{36,5}=0,3\left(mol\right)\)
a. \(V_{H_2\left(đktc\right)}=n.24,79=0,15.24,79=3,7185\left(l\right)\)
b. \(Fe_2O_3+3H_2\rightarrow^{t^0}2Fe+3H_2O\)
1 : 3 : 2 : 3 (mol)
0,075 : 0,15 (mol)
\(n_{Fe_2O_3}=\dfrac{m}{M}=\dfrac{12}{160}=0,075\left(mol\right)\)
-Chuyển thành tỉ lệ: \(\dfrac{0,075}{1}>\dfrac{0,15}{3}=0,05\)
\(\Rightarrow\)H2 phản ứng hết còn Fe2O3 dư.
\(Fe_2O_3+3H_2\rightarrow^{t^0}2Fe+3H_2O\)
1 : 3 : 2 : 3 (mol)
0,05 : 0,15 : 0,1 : 0,15 (mol)
\(\Rightarrow m_{Fe}=n.M=0,1.56=5,6\left(g\right)\)
nHCl=0,3(mol)
nZn=0,1(mol)
PTHH: Zn +2 HCl -> ZnCl2 + H2
ta có: 0,3/2 > 0,1/1
=> Zn hết, HCl dư, tính theo nZn.
=> nH2=nZn=0,1(mol)
=>V(H2,đktc)=0,1.22,4=2,24(l)
Chúc em học tốt!
$n_{Zn} = 0,1(mol) ; n_{HCl} = \dfrac{10,95}{36,5} = 0,3(mol)$
$Zn + 2HCl \to ZnCl_2 + H_2$
$n_{Zn} : 1 < n_{HCl} : 2$ nên Zn dư
$n_{H_2} = n_{Zn} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(l)$