Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,2 0,4 0,2 0,2
a) \(n_{H2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,2.22,4=4,48\left(l\right)\)
b) \(n_{ZnCl2}=\dfrac{0,2.1}{1}=0,2\left(mol\right)\)
⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)
c) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)
⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(C_{ddHCl}=\dfrac{14,6.100}{250}=5,84\)0/0
Chúc bạn học tốt
\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\)
Pt : \(2Al++3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2|\)
2 3 1 3
0,1 0,15 0,05 0,15
a) \(n_{H2}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,15.22,4=3,36\left(l\right)\)
b) \(n_{Al2\left(SO4\right)3}=\dfrac{0,15.1}{3}=0,05\left(mol\right)\)
⇒ \(m_{Al2\left(SO4\right)3}=0,05.342=17,1\left(g\right)\)
c) \(n_{H2SO4}=\dfrac{0,1.3}{2}=0,15\left(mol\right)\)
200ml = 0,2l
\(C_{M_{ddH2SO4}}=\dfrac{0,15}{0,2}=0,75\left(M\right)\)
Chúc bạn học tốt
\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)
\(PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
(mol)_____0,2____0,2______0,2____0,2__
\(a.V_{H_2}=22,4.0,2=4,48\left(l\right)\)
\(b.m_{ddH_2SO_4}=\dfrac{0,2.98.100}{24,5}=80\left(g\right)\)
\(c.m_{ddspu}=13+80-0,2.2=92,6\left(g\right)\\ \Rightarrow C\%_{ddspu}=\dfrac{0,2.136}{92,6}.100=29,4\left(\%\right)\)
`Fe + H_2 SO_4 -> FeSO_4 + H_2 ↑`
`0,3` `0,3` `0,3` `0,3` `(mol)`
`n_[Fe] = [ 16,8 ] / 56 = 0,3 (mol)`
`a) m_[dd H_2 SO_4] = [ 0,3 . 98 ] / [ 9,8 ] . 100 = 300 (g)`
`b) V_[H_2] = 0,3 . 22,4 = 6,72 (l)`
`c) C%_[FeSO_4] = [ 0,3 . 152 ] / [ 16,8 + 300 - 0,3 . 2 ] . 100 ~~ 14,42%`
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
Theo PT: \(n_{H_2SO_4}=n_{FeSO_4}=n_{H_2}=n_{Fe}=0,1\left(mol\right)\)
a, \(V_{H_2}=0,1.24,79=2,479\left(l\right)\)
b, \(m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
\(\Rightarrow m_{ddH_2SO_4}=\dfrac{9,8}{9,8\%}=100\left(g\right)\)
c, Ta có: m dd sau pư = 5,6 + 100 - 0,1.2 = 105,4 (g)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,1.152}{105,4}.100\%\approx14,42\%\)
a)nMgO=0,15(mol)
Ta có PTHH:
MgO+H2SO4->MgSO4+H2O
0,15......0,15...........0,15..................(mol)
Theo PTHH:mH2SO4=0,15.98=14,7g
b)Ta có:mddH2SO4=1,2.50=60(g)
=>Nồng độ % dd H2SO4là:
C%ddH2SO4=\(\dfrac{14,7}{60}100\)=24,5%
c)Theo PTHH:mMgSO4=0,15.120=18(g)
Khối lượng dd sau pư là:
mddsau=6+60=66(g)
Vậy nồng độ % dd sau pư là:
C%ddsau=\(\dfrac{18}{66}.100\)=27,27%
a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)
c, \(C\%_{H_2SO_4}=\dfrac{19,6}{50}.100\%=39,2\%\)
d, Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(a,n_{Zn}=\dfrac{1,3}{65}=0,02\left(mol\right)\)
PTHH: Zn + H2SO4 ---> ZnSO4 + H2
0,02--->0,02--------->0,02----->0,02
b, mZnSO4 = 0,02.161 = 3,22 (g)
c, VH2 = 0,02.22,4 = 0,448 (l)
d, \(m_{ddH_2SO_4}=\dfrac{0,02.98}{10\%}=19,6\left(g\right)\)
e, mdd = 19,6 + 1,3 - 0,02.2 = 20,86 (g)
=> \(C\%_{ZnSO_4}=\dfrac{0,02.161}{20,86}.100\%=15,44\%\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
Pt : \(Fe+H_2SO_4\rightarrow FeSO_4+H_2|\)
1 1 1 1
0,1 0,1 0,1 0,1
a) \(n_{H2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(V_{H2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)
b) \(n_{FeSO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
⇒ \(m_{FeSO4}=0,1.152=15,2\left(g\right)\)
c) \(n_{H2SO4}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)
\(m_{H2SO4}=0,1.98=9,8\left(g\right)\)
\(m_{ddH2SO4}=\dfrac{9,8.100}{10}=98\left(g\right)\)
Chúc bạn học tốt