Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(n_{BaCl_2}=\dfrac{200.10,4\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4+2HCl\\ b,Qu\text{ỳ}-t\text{í}m-ho\text{á}-\text{đ}\text{ỏ}-do-c\text{ó}-\text{ax}it-HCl\\ c,n_{BaSO_4}=n_{H_2SO_4}=n_{BaCl_2}=0,1\left(mol\right)\\ m_{k\text{ết}-t\text{ủa}}=m_{BaSO_4}=233.0,1=23,3\left(g\right)\\ d,m_{\text{dd}H_2SO_4}=\dfrac{0,1.98}{4,9\%}=200\left(g\right)\\ e,m_{\text{dd}HCl}=200+200-23,3=376,7\left(g\right)\\ n_{HCl}=0,1.2=0,2\left(mol\right)\\ C\%_{\text{dd}HCl}=\dfrac{0,2.36,5}{376,7}.100\approx1,938\%\)
\(n_{BaCl_2}=\dfrac{208.10\%}{208}=0,1\left(mol\right)\\ a,BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\\ 0,1............0,1..............0,1.............0,2\left(mol\right)\\ b,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{8}=122,5\left(g\right)\\ c,m_{kt}=m_{BaSO_4}=0,1.233=23,3\left(g\right)\\ d,m_{ddsau}=208+122,5-23,3=307,2\left(g\right)\\ C\%_{ddHCl}=\dfrac{0,2.36,5}{307,2}.100\approx2,376\%\)
Phương trình:
Fe3O4 + 4H2SO4 → FeSO4 + Fe2(SO4)3 + 4H2O
FeSO4 + 2NaOH → Na2SO4 + Fe(OH)2↓
Fe2(SO4)3 + 6NaOH → 3Na2SO4 + 2Fe(OH)3↓
2Fe(OH)2 + ½ O2 → Fe2O3 + 2H2O
2Fe(OH)3 → Fe2O3 + 3H2O
Fe2O3 + 3CO → 2Fe + 3CO2↑
CO2 + Ca(OH)2 → CaCO3↓ + H2O
3CO2 + Ca(OH)2 → Ca(HCO3)2
Ca(HCO3)2 → CaCO3 + CO2↑ + H2O
\(m_{H_2SO_4}=\dfrac{19,6\cdot20\%}{100\%}=3,92\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{3,92}{98}=0,04\left(mol\right)\\ PTHH:H_2SO_4+BaCl_2\rightarrow BaSO_4\downarrow+2HCl\\ \Rightarrow n_{H_2SO_4}=n_{BaCl_2}=n_{BaSO_4}=0,04\left(mol\right)\\ \Rightarrow m_{CT_{BaCl_2}}=0,04\cdot208=8,32\left(g\right)\\ \Rightarrow m_{dd_{BaCl_2}}=\dfrac{8,32\cdot100\%}{12\%}\approx69,3\left(g\right)\\ m_{kết.tủa}=m_{BaSO_4}=0,04\cdot233=9,32\left(g\right)\)
500ml=0.5l
nBaOH2 =0.5 x1=0.5 mol
MH2SO4=500.15%=75g
nH2SO4= xấp xỉ 0.8mol
H2SO4 dư tính theo BaOH2
pthh: Ba(OH)2 + H2SO4 => BaSO4+H2O
Theo pthh nBaSO4= nBa(OH)2=0.5mol
=>m kết tủa= 0.5x233=116.5g
theo pthh nH2SO4 phản ứng=nBaOH2= 0.5 mol
=> nH2SO4 Dư=0.8-0.5=0.3 mol
=>
m dư=0.3x98=29.4g
mH2SO4 đã dùng là m phản ứng? nếu thế thì m đã dung là 75-29.4=45.6
còn nếu m đã dùng là m chất tan thi là 75g như trên =))
\(n_{BaCl_2}=\dfrac{150.16,64\%}{137+35.2}=0,12\left(mol\right)\)
\(n_{H_2SO_4}=\dfrac{100.14,7\%}{98}=0,15\left(mol\right)\)
Phương trình hóa học :
BaCl2 + H2SO4 -----> BaSO4 + 2HCl
Dễ thấy \(\dfrac{n_{BaCl_2}}{1}< \dfrac{n_{H_2SO_4}}{1}\Rightarrow H_2SO_4\text{ dư }0,15-0,12=0,03\left(mol\right)\)
c) Khối lượng kết tủa :
\(m_{BaSO_4}=0,12.233=27,96\) (g)
Khối lượng chất tan : \(m_{HCl}=0,24.36,5=8,76\left(g\right)\) ;
\(m_{H_2SO_4\left(\text{dư}\right)}=0,03.98=2,94\left(g\right)\)
c) \(C\%_{H_2SO_4}\)= \(\dfrac{2,94}{150+100}.100\%=1,176\%\)
\(C\%_{HCl}=\dfrac{8,76}{150+100}.100\%=3.504\%\)
d) NaOH + HCl ---> NaCl + H2O
0,24 <-- 0,24
mol mol
2NaOH + H2SO4 ---> Na2SO4 + 2H2O
0,06 mol <-- 0,03 mol
\(\Rightarrow n_{NaOH}=0,24+0,06=0,3\left(mol\right)\)
\(V_{NaOH}=0,3.2=0,6\left(l\right)\)
\(m_{Fe_2\left(SO_4\right)_3}=\dfrac{200\cdot20}{100}=40\left(g\right)\Rightarrow n=0,1mol\)
\(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4\)
0,1 0,6 0,2 0,3
a)\(m_{NaOH}=0,6\cdot40=24\left(g\right)\)
b)\(m_{Fe\left(OH\right)_3}=0,2\cdot107=21,4\left(g\right)\)
c)\(m_{Na_2SO_4}=0,3\cdot142=42,6\left(g\right)\)
\(m_{ddsau}=200+24-21,4=202,6\left(g\right)\)
\(\Rightarrow C\%=\dfrac{42,6}{202,6}\cdot100\%=21,03\%\)
nNaOH = 0,1 mol
nH2SO4 = 0,1 mol
PT: 2NaOH + H2SO4 -> Na2SO4 + H2O
=> H2SO4 dư: 0,1 - 0,05= 0,05 (mol)
=> mH2SO4 dư = n. M = 0,05 . 98 = 4,9 g
a,\(2NaOH+H2SO4->Na2SO4+2H2O\)
b,theo pthh
PTHH:\(2NaOH+H2SO4->Na2SO4+2H2O\)
theo pthh:\(2\)..................1...........(mol)
theo bài: \(\dfrac{100}{1000}\)............\(\dfrac{9,8}{98}........\)(mol)
\(=>\dfrac{0,1}{2}< \dfrac{0,1}{1}\)=>H2SO4 dư
c,theo pthh \(=>nNA2SO4=\dfrac{1}{2}nNaOH=0,05mol\)
\(=>mNa2SO4=142.0,05=7,1g\)
a) \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
b) \(n_{BaCl_2}=\dfrac{52}{208}=0,25\left(mol\right)\)
PTHH: \(BaCl_2+H_2SO_4\rightarrow BaSO_4\downarrow+2HCl\)
0,25----->0,25------->0,25---->0,5
=> \(m_{H_2SO_4}=0,25.98=24,5\left(g\right)\)
=> \(m_{ddH_2SO_4}=\dfrac{24,5.100}{19,6}=125\left(g\right)\)
c) \(m_{BaSO_4}=0,25.233=58,25\left(g\right)\)
d)
\(n_{NaOH}=\dfrac{8}{40}=0,2\left(mol\right)\)
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
Xét tỉ lệ \(\dfrac{0,5}{1}>\dfrac{0,2}{1}\) => NaOH hết, HCl dư
=> Quỳ tím chuyển màu đỏ