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\(n_{SO_3}=\dfrac{200}{80}=2,5\left(mol\right)\)
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)
\(C\%_{dd.H_2SO_4.sau.pư}=\dfrac{435,4}{1320}.100\%=32,985\%\)
\(n_{SO_3}=\dfrac{200}{80}=2.5\left(mol\right)\)
\(m_{dd_{H_2SO_4}}=1000\cdot1.12=1120\left(g\right)\)
\(m_{H_2SO_4}=1120\cdot17\%=190.4\left(g\right)\)
\(SO_3+H_2O\rightarrow H_2SO_4\)
\(2.5.....................2.5\)
\(m_{H_2SO_4}=2.5\cdot98+190.4=435.4\left(g\right)\)
\(m_{dd_{H_2SO_4}}=200+1120=1320\left(g\right)\)
\(C\%_{H_2SO_4}=\dfrac{435.4}{1320}\cdot100\%=33\%\)
$SO_3 + H_2O \to H_2SO_4$
$n_{H_2SO_4} = n_{SO_3} = \dfrac{200}{80} = 2,5(mol)$
$m_{dd\ H_2SO_4\ 17\%} =1000.1,12 = 1120(gam)$
Sau khi pha :
$m_{dd} = 200 + 1120 = 1320(gam)$
$m_{H_2SO_4} = 1120.17\% + 2,5.98 = 435,4(gam)$
$C\%_{H_2SO_4} = \dfrac{435,4}{1320}.100\% = 33\%$
\(SO_3+H_2O\rightarrow H_2SO_4\\ n_{SO_3}=a\left(mol\right)\\ \rightarrow m_{SO_3}=80a\left(g\right);m_{H_2SO_4}=98a\left(g\right)\\ Vì:dd.thu.được.nồng.độ.20\%,nên.ta.có:\\ \dfrac{200.14,7\%+98a}{80a+200}.100\%=20\%\\ \Leftrightarrow a=12,927\\ Vậy:m=m_{SO_3}=12,927\left(g\right)\)
mdung dịch H2SO4 = 500.1,2 =600(g)
=> mH2SO4 = 600.24,5% =147(g)
PTHH : SO3 + H2O -> H2SO4
nSO3 =\(\dfrac{m}{80}\) mol
Theo PTHH => nH2SO4 = m/80 (mol) => mH2SO4 = 1,225m(g)
=> 147 (g)
mdung dịch = m + 600 (g)
Theo bài ra ta có hệ :
\(\dfrac{1,225m+147}{m+600}\).100%=49%
=>m=200g
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%
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mdd H2SO4 = 1,12 . 500 = 560 (g)
mH2SO4 = \(\dfrac{17\times560}{100}=95,2\left(g\right)\)
=> mH2O = 560 - 95,2 = 464,8 (g)
=> nH2O = \(\dfrac{464,8}{18}=25,82\left(mol\right)\)
nSO3 = \(\dfrac{100}{80}=1,25\left(mol\right)\)
=> H2O dư
Pt: SO3 + H2O --> H2SO4
1,25 mol----------> 1,25 mol
mH2SO4 = 1,25 . 98 = 122,5 (g)
mH2SO4 sau khi trộn = 122,5 + 95,2 = 217,7 (g)
mdd = 560 + 100 = 660 (g)
C% = \(\dfrac{217,7}{660}.100\%=33\%\)
BT1:
\(m_{NaCl}=50.20\%+150.10\%=25\left(g\right)\)
\(m_{ddNaCl}=50+150=200\left(g\right)\)
\(C\%_{ddNaCl}=\dfrac{25.100\%}{200}=12,5\%\)
BT2:
\(n_{H_2SO_4}=0,2.5+0,2.3=1,6\left(mol\right)\)
\(V_{ddH_2SO_4}=0,2+0,2=0,4\left(l\right)\)
\(C_{M_{ddH_2SO_4}}=\dfrac{1,6}{0,4}=4M\)
SO3 + H2O → H2SO4
\(m_{ddH_2SO_4.17\%}=1000\times1,12=1120\left(g\right)\)
\(\Rightarrow m_{H_2SO_4.17\%}=1120\times17\%=190,4\left(g\right)\)
\(n_{SO_3}=\dfrac{200}{80}=2,5\left(mol\right)\)
Theo PT: \(n_{H_2SO_4}tt=n_{SO_3}=2,5\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}tt=2,5\times98=245\left(g\right)\)
\(\Rightarrow m_{H_2SO_4}mới=m_{H_2SO_4}tt+m_{H_2SO_4.17\%}=245+190,4=435,4\left(g\right)\)
\(m_{dd}mới=m_{SO_3}+m_{ddH_2SO_4.17\%}=200+1120=1320\left(g\right)\)
\(\Rightarrow C\%_{dd}mới=\dfrac{m_{H_2SO_4}mới}{m_{dd}mới}=\dfrac{435,4}{1320}\times100\%=32,98\%\)
PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\\ 2,5mol:2,5mol\rightarrow2,5mol\)
\(n_{SO_3}=\dfrac{200}{80}=2,5\left(mol\right)\)
\(m_{H_2SO_4}=2,5.98=245\left(g\right)\)
\(m_{ddH_2SO_417\%}=1000.1,12=1120\left(g\right)\)
\(m_{H_2SO_4trongdd}=17\%.1120=190,4\left(g\right)\)
\(C\%dd=\dfrac{245+190,4}{1120+200}.100\%=32,98\%\)