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PTHH: \(BaCl_2+H_2SO_4\rightarrow2HCl+BaSO_4\downarrow\)
a+b) Ta có: \(n_{BaCl_2}=\dfrac{400\cdot5,2\%}{208}=0,1\left(mol\right)=n_{H_2SO_4}=n_{BaSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,1\cdot98}{20\%}=49\left(g\right)\\m_{BaSO_4}=0,1\cdot233=23,3\left(g\right)\end{matrix}\right.\)
c) Theo PTHH: \(n_{HCl}=0,2\left(mol\right)\) \(\Rightarrow m_{HCl}=0,2\cdot36,5=7,3\left(g\right)\)
Mặt khác: \(m_{dd\left(sau.p/ứ\right)}=m_{ddBaCl_2}+m_{ddH_2SO_4}-m_{BaSO_4}=425,7\left(g\right)\)
\(\Rightarrow C\%_{HCl}=\dfrac{7,3}{425,7}\cdot100\%\approx1,71\%\)
Bạn xem lại giúp mình , coi đề có bị thiếu gì không nhé
Bài 19 :
\(a) n_{Al} = \dfrac{10,8}{27} = 0,4(mol)\\ 2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\\ n_{H_2} = \dfrac{3}{2}n_{Al} = 0,6(mol)\\ V_{H_2} = 0,6.22,4 = 13,44(lít)\\ b) \text{Chất tan : }Al_2(SO_4)_3\\ n_{Al_2(SO_4)_3} = \dfrac{1}{2}n_{Al} = 0,2(mol)\\ m_{Al_2(SO_4)_3} = 0,2.342 = 68,4(gam)\)
Bài 18 :
\(a) n_{HCl} = \dfrac{250.7,3\%}{36,5 } = 0,5(mol)\\ Zn + 2HCl \to ZnCl_2 + H_2\\ n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,25(mol) \Rightarrow V_{H_2} = 0,25.22,4 = 5,6(lít)\\ b) \text{Chất tan : } ZnCl_2\\ n_{ZnCl_2} = n_{H_2} = 0,25(mol)\\ m_{ZnCl_2} = 0,25.136 = 34(gam)\)
\(m_{HCl}=\dfrac{300.7,3\%}{100\%}=21,9g\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6mol\\ HCl+NaOH\rightarrow NaCl+H_2O\left(1\right)\\ n_{NaOH\left(1\right)}=n_{HCl}=0,6mol\\ m_{H_2SO_4}=\dfrac{200.9,8\%}{100\%}=19,6g\\ n_{H_2SO_4}=\dfrac{19,6}{98}=0,2mol\\ H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\left(2\right)\\ n_{NaOH\left(2\right)}=0,2.2=0,4mol\\ n_{NaOH}=0,4+0,6=1mol\\ m_{NaOH}=1.40=40g\\ m_{ddNaOH}=\dfrac{40}{5\%}\cdot100\%=800g\)
mNa2CO3=100*19.96/100=19.96g
nNa2CO3=19.96/106=0.19mol
mBaCl2=200*10.04/100=20.08g
nBaCl2=20.08/208=0.097mol
Na2CO3 + BaCl2 -> BaCO3 + 2NaCl
(mol) 1 1
(mol) 0.19 0.097
Lập tỉ lệ: 0.19> 0.097. Na2Co3 dư dư
Na2CO3 + BaCl2 -> BaCO3 + 2NaCl
(mol) 0.097 0.097 0.097 0.194
mdd = mddNa2CO3 + mddBaCl2 - mBaCO3
=100+200-0.097*197=208.891g
nNa2CO3 dư = 0.19-0.097=0.093mol
mNa2CO3 dư = 0.093*106=9.858g
mNaCl = 0.194*58.5=11.349g
C%Na2CO3 dư = 9.858/208.891*100=4.72%
C%NaCl = 11.349/208.891*100=5.43%
\(m_{dd.BaCl_2}=400.1,003=401,2\left(g\right)\)
=> \(n_{BaCl_2}=\dfrac{401,2.5,2\%}{208}=0,1003\left(mol\right)\)
\(m_{dd.H_2SO_4}=100.1,14=114\left(g\right)\)
=> \(n_{H_2SO_4}=\dfrac{114.20\%}{98}=\dfrac{57}{245}\left(mol\right)\)
PTHH: BaCl2 + H2SO4 --> BaSO4 + 2HCl
Xét tỉ lệ: \(\dfrac{0,1003}{1}< \dfrac{\dfrac{57}{245}}{1}\) => BaCl2 hết, H2SO4 dư
PTHH: BaCl2 + H2SO4 --> BaSO4 + 2HCl
0,1003->0,1003-->0,1003-->0,2006
mdd sau pư = 401,2 + 114 - 0,1003.233 = 491,8301 (g)
\(\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{98\left(\dfrac{57}{245}-0,1003\right)}{491,8301}.100\%=2,637\%\\C\%_{HCl}=\dfrac{0,2006.36,5}{491,8301}.100\%=1,489\%\end{matrix}\right.\)
mddBaCl2 = 1,003 . 400 = 401,2 (g) mBaCl2 = 401,2 . 5,2% = 20,8624 (g)
nBaCl2 = 20,8624/208 = 0,1003 (mol)
mddH2SO4 = 1,14.100 = 114 (g) mH2SO4 = 114 . 20% = 22,8 (g)
nH2SO4 = 22,8/98 (mol)
PTHH: BaCl2 + H2SO4 -> BaSO4 + 2HCl
Bđ: 0,1003 22,8/98
Pư: 0,1003 -> 0,1003 -> 0,1003 -> 0,2006 (mol)
Sau: 0 0,132 0,1003 0,2006 (mol)
Dung dịch sau phản ứng chứa:
mH2SO4 dư = 22,8 - 0,1003.98 = 12,9706 (g)
mHCl = 0,2006.36,5 = 7,3219 (g)
Khối lượng dd sau pư: mdd sau pư = mddBaCl2 + mddH2SO4 - mBaSO4
= 401,2 + 114 - 0,1003.233 = 491,8301 (g)
Nồng độ phần trăm:
C% H2SO4 = (12,9706/491,8301).100% ≈ 2,64%
C% HCl = (7,3219/491,8301).100% ≈ 1,49%
`n_{BaCl_2}={20,8}/{208}=0,1(mol)`
`n_{H_2SO_4}={20.19,6\%}/{98}=0,04(mol)`
`H_2SO_4+BaCl_2->BaSO_4+2HCl`
`0,04->0,04->0,04->0,08(mol)`
Do `0,1>0,04->BaCl_2` dư.
`C\%_{BaCl_2\ du}={208(0,1-0,04)}/{20,8+20-0,04.233}.100\%\approx 39,64\%`
`C\%_{HCl}={0,08.36,5}/{20,8+20-0,04.233}.100\%\approx 9,28\%`
Ta có: \(m_{BaCl_2}=200.5,2\%=10,4\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{10,4}{208}=0,05\left(mol\right)\)
\(m_{H_2SO_4}=58,8.20\%=11,76\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{11,76}{98}=0,12\left(mol\right)\)
PT: \(BaCl_2+H_2SO_4\rightarrow BaSO_{4\downarrow}+2HCl\)
Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,12}{1}\), ta được H2SO4 dư.
Theo PT: \(\left\{{}\begin{matrix}n_{H_2SO_4\left(pư\right)}=n_{BaSO_4}=n_{BaCl_2}=0,05\left(mol\right)\\n_{HCl}=2n_{BaCl_2}=0,1\left(mol\right)\end{matrix}\right.\)
⇒ nH2SO4 (dư) = 0,12 - 0,05 = 0,07 (mol)
Ta có: m dd sau pư = m dd BaCl2 + m dd H2SO4 - mBaSO4 = 200 + 58,8 - 0,05.233 = 247,15 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,07.98}{247,15}.100\%\approx2,78\%\\C\%_{HCl}=\dfrac{0,1.36,5}{247,15}.100\%\approx1,48\%\end{matrix}\right.\)
Bạn tham khảo nhé!