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%
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 = 32.985%
