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nK = 39 / 39=1 (mol)
Pt: 2K + 2H2O --> 2KOH + H2
1 mol--------------------------> 0,5 mol
mH2 = 0,5 . 2 = 1 (g)
mdd = mK + mnước - mH2 = 39 + 362 - 1 = 400 (g)
C% dd KOH = 39/400.100%=9,75%
https://hoc24.vn/hoi-dap/tim-kiem?id=562460&q=t%C3%ADnh%20n%E1%BB%93ng%20%C4%91%E1%BB%99%20ph%E1%BA%A7n%20tr%C4%83m%20c%E1%BB%A7a%20dung%20d%E1%BB%8Bch%20t%E1%BB%8Da%20th%C3%A0nh%20khi%20h%C3%B2a%20tan%20%3A%20%201%2F%2039g%20Kali%20v%C3%A0o%20362g%20n%C6%B0%E1%BB%9Bc%20%202%2F200g%20So3%20v%C3%A0o%201%20l%C3%ADt%20dung%20d%E1%BB%8Bch%20H2SO4%2017%25%20%28D%20%3D%201%2C12%20G%2FML%29
\(n_{SO_3}=\dfrac{200}{80}=2,5\left(mol\right)\)
mdd H2SO4 17% = 1000.1,12 = 1120 (g)
=> \(m_{H_2SO_4}=\dfrac{1120.17}{100}=190,4\left(g\right)\)
PTHH: SO3 + H2O --> H2SO4
2,5------------>2,5
=> mH2SO4(sau pư) = 2,5.98 + 190,4 = 435,4 (g)
mdd sau pư = 200 + 1120 = 1320 (g)
\(C\%_{dd.H_2SO_4.sau.pư}=\dfrac{435,4}{1320}.100\%=32,985\%\)
SO3 + H2O => H2SO4
nSO3 = m/M = 200/80 = 2.5 (mol)
Theo phương trình: mH2SO4 = n.M = 98x2.5 = 245g
V = 1l=1000 ml, D =1.12g/ml
mddH2SO4 17% = D.V = 1000x1.12 = 1120g
mH2SO4 = 1120x17/100 = 190.4 (g)
C% = (190.4+245)x100/1365 = 31.9%
SO3 + H2O---> H2SO4
nSO3=200/80=2,5(mol)
Theo pt:
nSO3=nH2SO4=2,5(mol)
mH2SO4=98.2,5=245(g)
mdd H2SO4 17 % =1000.1,12=1120(g)
mH2SO4trong dd=1120.17/100=190,4(g)
=>C%
SO3 + H2O => H2SO4
nSO3 = m/M = 200/80 = 2.5 (mol)
Theo phương trình: mH2SO4 = n.M = 98x2.5 = 245g
V = 1l=1000 ml, D =1.12g/ml
mddH2SO4 17% = D.V = 1000x1.12 = 1120g
mH2SO4 = 1120x17/100 = 190.4 (g)
C% = (190.4+245)x100/1365 = 31.9%
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(n_{SO_3}=\dfrac{12}{80}=0,15\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}=n_{SO_3}=0,15\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,15.98}{12+100}.100\%=13,125\%\)
a, PTPƯ: SO3 + H2O ---> H2SO4
nSO3=\(\dfrac{2,24}{22,4}=0,1mol\)
1 mol SO3 ---> 0,1 mol H2SO4
nên 0,1 mol SO3 ---> 0,1 mol H2SO4
CM H2SO4=\(\dfrac{0,1}{0,5}\)=0,2 M
b, PTPƯ: Zn + H2SO4 ---> ZnSO4 + H2
1 mol H2SO4 ---> 1 mol Zn
nên 0,1 mol H2SO4 ---> 0,1 mol Zn
mZn=0,1.65=6,5 g
Ta có: \(m_{ddH_2SO_4\left(60\%\right)}=700.1,503=1052,1\left(g\right)\Rightarrow m_{H_2SO_4}=1052,1.60\%=631,26\left(g\right)\)
\(m_{ddH_2SO_4\left(20\%\right)}=500.1,1476=573,8\left(g\right)\Rightarrow m_{H_2SO_4}=573,8.20\%=114,76\left(g\right)\)
ΣmH2SO4 = 631,26 + 114,76 = 746,02 (g)
\(n_{H_2}=\dfrac{1,792}{22,4}=0,08\left(mol\right)\)
PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)
Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,08\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,08.98=7,84\left(g\right)\)
\(\Rightarrow\dfrac{746,02}{V}=\dfrac{7,84}{0,2}\Rightarrow V\approx19,03\left(l\right)\)
\(n_{H_2SO_4}\left(3M\right)=2,1mol\)
\(n_{H_2SO_4}\left(6M\right)=3mol\)
\(\rightarrow CM_{H_2SO_4}=\frac{2,1+3}{0,7+0,5}=4,25M\)
a)nMgO=6:40=0,15(mol)
Ta có PTHH:
MgO+H2SO4->MgSO4H2O
0,15......0,15...........0,15..................(mol)
Theo PTHH:mH2SO4=0,15.98=14,7g
b)Ta có:mddH2SO4=D.V=1,2.50=60(g)
=>Nồng độ % dd H2SO4 là:
C%ddH2SO414,7\60.100%=24,5%
c)Theo PTHH:mMgSO4=0,15.120=18(g)
Khối lượng dd sau pư là:
mddsau=mMgO+mddH2SO44=6+60=66(g)
Vậy nồng độ % dd sau pư là:
C%ddsau=18\66.100%=27,27%
\(n_{SO_3}=\dfrac{100}{80}=1,25\left(mol\right)\\ n_{H_2SO_4\left(bđ\right)}=1.3=3\left(mol\right)\\ m_{H_2SO_4\left(bđ\right)}=3.98=294\left(g\right)\\ m_{ddH_2SO_4\left(bđ\right)}=1000.1.1,12=1120\left(g\right)\\ m_{ddH_2SO_4\left(sau\right)}=1120+100=1220\left(g\right)\)
PTHH: SO3 + H2O ---> H2SO4
1,25-------------->1,25
=> \(m_{H_2SO_4\left(sau\right)}=294+1,25.98=416,5\left(g\right)\)
=> \(C\%_{H_2SO_4\left(sau\right)}=\dfrac{416,5}{1220}.100\%=34,14\%\)
hình như mik đọc lộn đề r:VVV