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a, \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b, \(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
c, \(n_{ZnCl_2}=n_{Zn}=0,2\left(mol\right)\Rightarrow m_{ZnCl_2}=0,2.136=27,2\left(g\right)\)
d, \(n_{HCl}=2n_{Zn}=0,4\left(mol\right)\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\dfrac{14,6}{7,3\%}=200\left(g\right)\)
⇒ m dd sau pư = 13 + 200 - 0,2.2 = 212,6 (g)
\(\Rightarrow C\%_{ZnCl_2}=\dfrac{27,2}{212,6}.100\%\approx12,79\%\)
a, \(H_2SO_4+Zn=ZnSO_4+H_2\uparrow\)
b,
\(n_{Zn}=\frac{13}{65}=0,2\left(mol\right)\)
Theo PTHH : \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow V_{H_2=}=n_{H_2}\cdot22,4=0,2\cdot22,4=4,48\left(l\right)\)
nZn = 0,1 mol
Zn + 2HCl \(\rightarrow\) ZnCl2 + H2
\(\Rightarrow\) V = \(\dfrac{0,1}{2}\) = 0,05 (l)
\(\Rightarrow\) VH2 = 0,1.22,4 = 2,24 (l)
\(\Rightarrow\) mZnCl2 = 0,1.136 = 13,6 (g)
Zn + 2 HCl -> ZnCl2 + H2
1mol 2mol 1mol 1mol
0,1 mol
số mol của Zn
nZn = 6,5/ 65 = 0,1 mol
dựa vào pthh ta có số mol của HCl
nHCl = 0,1 *2 /1 = 0,2 mol
VHCl = 0,2 /2 = o,1 lít
c. thiếu thiếu .....
\(a,2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{HCl}=0,2.1,5=0,6\left(mol\right)\\ n_{H_2}=\dfrac{3}{6}.0,6=0,3\left(mol\right);n_{Al}=\dfrac{2}{6}.0,6=0,2\left(mol\right)\\ b,V=V_{H_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\\ m=m_{Al}=0,2.27=5,4\left(g\right)\)
nZn=0,1 mol
Zn +2HCl=> ZnCl2+ H2
0,1 mol =>0,2 mol
=>mHCl=36,5.0,2=7,3g
=>m dd HCl=7,3/14,6%=50g
mdd sau pứ=6,5+50-0,1.2=56,3g
=>C% dd ZnCl2=(0,1.136)/56,3.100%=24,16%
a.b. Zn + 2HCl ---> ZnCl2 + H2 (1)
Theo pt: 65g 73g 136g 2g
Theo đề: 6,5g 7,3g 13,6g
=> mddHCl=\(\frac{7,3.100}{14,6}=50\left(g\right)\)
c. Từ pt (1), ta có: \(C_{\%}=\frac{13,6}{50+6,5}.100\%=24,1\%\)
PTHH: Zn + 2HCl => ZnCl2 + H2
=> nHCl =2nZn= 6,3/65 .2=63/325 (mol)
=> V= (63/325)/2 = 63/650 \(\approx\) 0,1 lít
=> nH2=nZn=6,3/65 (mol)
=> VH2=22,4.(6,3/65) xấp xỉ 2,17 lít
=> nZnCl2 =nZn=6,3/65 (mol)
=> m ZnCl2 = (65+35,5.2)(6,3/65) \(\approx\)13,18 g
a) PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
b) Ta có: \(n_{Zn}=\frac{6,3}{65}=\frac{63}{650}\left(mol\right)\) \(\Rightarrow n_{HCl}=\frac{63}{650}\cdot2=\frac{63}{325}\left(mol\right)\)
\(\Rightarrow V_{HCl}=\frac{2}{\frac{63}{325}}\approx10,3\left(l\right)\)
c) Theo PTHH: \(n_{Zn}=n_{H_2}=\frac{63}{650}\left(mol\right)\) \(\Rightarrow V_{H_2}=22,4\cdot\frac{63}{650}\approx2,2\left(l\right)\)
d) Theo PTHH: \(n_{Zn}=n_{ZnCl_2}=\frac{63}{650}\left(mol\right)\)
\(\Rightarrow m_{ZnCl_2}=\frac{63}{650}\cdot136\approx13,2\left(g\right)\)
a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$
b) n Al = 8,1/27 = 0,3(mol)
Theo PTHH :
n H2 = 3/2 n Al = 0,45(mol)
V H2 = 0,45.22,4 = 10,08(lít)
c) n AlCl3 = n Al = 0,3(mol)
m AlCl3 = 0,3.133,5 = 40,05(gam)
d) n HCl = 3n Al = 0,9(mol)
m dd HCl = 0,9.36,5/7,3% = 450(gam)
Sau phản ứng :
m dd = 8,1 + 450 -0,45.2 = 457,2(gam)
C% AlCl3 = 40,05/457,2 .100% = 8,76%
a) PTHH: Zn(0,1) + 2HCl(0,2) -> ZnCl2(0,1) + H2(0,1)
b) nZn=6,5:65=0,1(mol)
Theo pt ý a) ta có nHCl=0,2(mol)
-> VHCl=0,2.22,4=4,48(l)
c) Theo pt ý a) ta có: nH2=0,1(mol)
-> VH2=0,1.22,4=2,24(l)
d) Theo pt ý a) ta có: nZnCl2= 0,1(mol)
-> mZnCl2=0,1.136=13,1(g)
Câu d) m\(_{ZnCl_{ }2}\)\(_{ }\)= 0,1.136=13,6(g)