\(n_{SO_3}=\dfrac{200}{80}=2,5\left(mol\right)\)

m_{dd 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

=> m_{H2SO4(sau pư)} = 2,5.98 + 190,4 = 435,4 (g)

m_{dd sau pư} = 200 + 1120 = 1320 (g)

\(C\%_{dd.H_2SO_4.sau.pư}=\dfrac{435,4}{1320}.100\%=32,985\%\)

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: \(SO_3+H_2O\rightarrow H_2SO_4\\ 2,5mol:2,5mol\rightarrow2,5mol\)

\(n_{SO_3}=\dfrac{200}{80}=2,5\left(mol\right)\)

\(m_{H_2SO_4}=2,5.98=245\left(g\right)\)

\(m_{ddH_2SO_417\%}=1000.1,12=1120\left(g\right)\)

\(m_{H_2SO_4trongdd}=17\%.1120=190,4\left(g\right)\)

\(C\%dd=\dfrac{245+190,4}{1120+200}.100\%=32,98\%\)

SO3 + H2O -> H2SO4
(32+3*16)= 80........(2+32+4*16)=98
200g.......................x(g)
x= (200 * 98) / 80 = 245g
khoi luong dd H2SO4 truoc phan ung la: m = d * v = 1.12 * 1000 = 1120g
khoi luong H2SO4 truoc pu la: m = (1120 * 17) / 100 = 190.4g
khoi luong H2SO4 sau pu la: m = 190.4 + 245 = 435.4g
khoi luong dd H2SO4 sau pu la: m = 1120 + 200 = 1320g
C% cua dd thu duoc la: C%= ( 435.4 / 1320) *100 = 32.98%

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

0,2------->0,1--------->0,1

Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư

Trong dung dịch D có:

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

b

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,35<---------0,35

\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)

2

b

mNaCl=\(\dfrac{200.15}{100}\)=30(g)

nNaCl=\(\dfrac{30}{58,5}\)=0.51(mol)

VddNaCl=\(\dfrac{200}{1,1}\)=181.8(ml)=0.1818(l)

CMNaCl=\(\dfrac{0,51}{0,1818}\)=2.8(M)

Câu B chịu nka~~

→nH2O(dd bđ)=1120−190,418=51,6(mol)→nH2O(dd bđ)=1120−190,418=51,6(mol)

$S{}^{}+{}^{}O\to {}^{}S{}^{}$

$\to S{}^{}$ hết, ${}^{}O$ dư

nH2

PTHH: \(SO_3+H_2O\rightarrow H_2SO_4\)

Ta có: \(n_{SO_3}=\dfrac{200}{80}=2,5\left(mol\right)=n_{H_2SO_4}\)

Mặt khác: \(m_{ddH_2SO_4}=1000\cdot1,12=1120\left(g\right)\) \(\Rightarrow m_{H_2SO_4\left(ban.đầu\right)}=1120\cdot17\%=190,4\left(g\right)\)

Bảo toàn khối lượng: \(m_{ddH_2SO_4\left(sau\right)}=m_{SO_3}+m_{ddH_2SO_4\left(ban.đầu\right)}=1320\left(g\right)\)

\(\Rightarrow C\%=\dfrac{2,5\cdot98+190,4}{1320}\cdot100\%\approx32,98\%\)

n_K = 39/400.100%=9,75%

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