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\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)
PTHH :
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,4 1,2 0,4 0,6
\(a,V_{H_2}=0,6.22,4=13,44\left(l\right)\)
\(b,m_{HCl}=1,2.36,5=43,8\left(g\right)\)
\(c,m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\)
\(m_{ddHCl}=\dfrac{43,8.100}{10}=438\left(g\right)\)
\(m_{ddAlCl_3}=10,8+438-\left(0,6.2\right)=447,6\left(g\right)\)
\(C\%=\dfrac{53,8}{447,6}.100\%\approx12,02\%\)
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: \(n_{H_2}=n_{Cu}=0,1\left(mol\right)\)
\(\Rightarrow m_{Cu}=0,1\cdot64=6,4\left(g\right)\)
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\)
\(m_{HCl}=\dfrac{150.18,25}{100}=27,375\left(g\right)\)
\(n_{HCl}=\dfrac{27,375}{36,5}=0,75\left(mol\right)\)
PTHH :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
trc p/u : 0,3 0,75
p/u: 0,3 0,6 0,3 0,3
sau p/u : 0 0,15 0,3 0,3
---> Sau p/ư HCl dư
\(a,V_{H_2}=0,3.22,4=6,72\left(l\right)\)
\(b,m_{ddHCl}=0,6.36,5=21,9\left(g\right)\)
\(c,m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)
\(m_{ddZnCl_2}=19,5+150-\left(0,3.2\right)=168,9\left(g\right)\)
\(C\%=\dfrac{40,8}{168,9}.100\%\approx24,16\%\)
1/ nMgO= 16/40=0.4 (mol)
MgO + 2HCl --> MgCl2 + H2
Từ PTHH:
nMgCl2= 0.4 (mol)
mMgCl2= 0.4*95=38g
nHCl= 0.8 (mol)
VHCl= 0.8/0.5=1.6 (l)
2/ Đặt: nAl= x (mol), nFe= y (mol)
mhh= 27x + 56y= 11g (1)
2Al + 3H2SO4 --> Al2(SO4)3 + 3H2
Fe + H2SO4 --> FeSO4 + H2
Từ PTHH:
nH2= 1.5x + y= 8.96/22.4=0.4 (mol) (2)
Giải (1) và (2):
x=0.2
y=0.1
mAl= 0.2*27=5.4g
%Al= 5.4/11*100%= 49.09%
%Fe= 5.6/11*100%= 50.91%
nH2SO4= 0.3+0.1=0.4 (mol)
mH2SO4= 0.4*98=39.2g
mddH2SO4= 39.2*100/20=196g
mdd sau phản ứng= 11+196-0.8=206.2g
C%Al2(SO4)3= 34.2/206.2*100=16.58%
C%FeSO4= 15.2/206.2*100= 7.37%
nAl = 5.4 / 27 = 0.2 (mol)
2Al + 6HCl => 2AlCl3 + 3H2
0.2......0.6............0.2.......0.3
a) VH2 = 0.3 * 22.4 = 6.72 (l)
b) mAlCl3 = 0.2 * 133.5 = 26.7 (g)
c) VddHCl = 0.6 / 1.5 = 0.4 (l)
d) CMAlCl3 = 0.2 / 0.4 = 0.5 (M)
PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,6\left(mol\right)\\n_{AlCl_3}=0,2\left(mol\right)\\n_{H_2}=0,3\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0,3\cdot22,4=6,72\left(l\right)\\m_{AlCl_3}=0,2\cdot133,5=26,7\left(g\right)\\V_{HCl}=\dfrac{0,6}{1,5}=0,4\left(l\right)=400\left(ml\right)\\C_{M_{AlCl_3}}=\dfrac{0,2}{0,4}=0,5\left(M\right)\end{matrix}\right.\)
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
\(m_{HCl}=\dfrac{300.10,95}{100}=32,85\left(g\right)\)
\(n_{HCl}=\dfrac{32,85}{36,5}=0,9\left(mol\right)\)
PTHH :
\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)
trc p/u: 0,4 0,9
p/u : 0,4 0,8 0,4 0,4
sau p/u: 0 0,1 0,4 0,4
---> sau p/ư : HCl dư
\(a,V_{H_2}=0,4.22,4=8,96\left(l\right)\)
\(b,m_{HCl}=0,8.36,5=29,2\left(g\right)\)
\(c,m_{ZnCl_2}=0,4.136=54,4\left(g\right)\)
\(m_{ddZnCl_2}=26+300-\left(0,4.2\right)=325,2\left(g\right)\)
\(C\%_{ZnCl_2}=\dfrac{54,4}{325,2}.100\%\approx16,73\%\)
\(m_{HCldư}=0,1.36,5=3,65\left(g\right)\)
\(C\%_{HCldư}=\dfrac{3,65}{300}.100\%\approx1,22\%\)
mình không chắc là có phải tính lượng dư của HCl không nên câu nãy bạn cứ tính tương tự cho chắc ăn nha
a. Ta có: mNaOH=\(\frac{200.20}{100}=40\left(g\right)\)
pt : NaOH + HCl --------> NaCl + H2O
theo pt: 40g 36,5g 58,5g 18g
theo đề: 40g 36,5g 58,5g
=>\(C_{\%}=\frac{58,5}{200+100}.100\%=19,5\%\)
b.\(C_{\%}=\frac{36,5}{100}.100\%=36,5\%\)
a) $2Al + 6HCl \to 2AlCl_3 + 3H_2$
n Al = 5,4/27 = 0,2(mol)
Theo PTHH : n H2 = 3/2 n Al = 0,3(mol)
=> V H2 = 0,3.22,4 = 6,72(lít)
b) n HCl = 3n Al = 0,6(mol)
=> mdd HCl = 0,6.36,5/20% = 109,5 gam
c)Sau phản ứng,
mdd = m Al + mdd HCl - m H2 = 5,4 + 109,5 - 0,3.2 = 114,3(gam)
=> C% AlCl3 = 0,2.133,5/114,3 .100% = 23,36%
cảm ơn bn nha