Câu hỏi của Shizadon

Question

Khi sục 200g SO3 vào 1 lít dd H2SO4 17% ( D = 1,12 g/ml). Tính C% của dd thu đc

Phùng Hà Châu · Accepted Answer

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: $n_{H_2SO_4}tt=n_{SO_3}=2,5\left(mol\right)$

$\Rightarrow m_{H_2SO_4}tt=2,5\times98=245\left(g\right)$

$\Rightarrow m_{H_2SO_4}mới=m_{H_2SO_4}tt+m_{H_2SO_4.17\%}=245+190,4=435,4\left(g\right)$

$m_{dd}mới=m_{SO_3}+m_{ddH_2SO_4.17\%}=200+1120=1320\left(g\right)$

$\Rightarrow C\%_{dd}mới=\dfrac{m_{H_2SO_4}mới}{m_{dd}mới}=\dfrac{435,4}{1320}\times100\%=32,98\%$Câu trả lời: 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: $n_{H_2SO_4}tt=n_{SO_3}=2,5\left(mol\right)$

$\Rightarrow m_{H_2SO_4}tt=2,5\times98=245\left(g\right)$

$\Rightarrow m_{H_2SO_4}mới=m_{H_2SO_4}tt+m_{H_2SO_4.17\%}=245+190,4=435,4\left(g\right)$

$m_{dd}mới=m_{SO_3}+m_{ddH_2SO_4.17\%}=200+1120=1320\left(g\right)$

$\Rightarrow C\%_{dd}mới=\dfrac{m_{H_2SO_4}mới}{m_{dd}mới}=\dfrac{435,4}{1320}\times100\%=32,98\%$

Thục Trinh · Answer

PTHH: $SO_3+H_2O\rightarrow H_2SO_4\
2,5mol:2,5mol\rightarrow2,5mol$

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

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

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

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

$C\%dd=\dfrac{245+190,4}{1120+200}.100\%=32,98\%$Câu trả lời: PTHH: $SO_3+H_2O\rightarrow H_2SO_4\
2,5mol:2,5mol\rightarrow2,5mol$

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

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

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

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

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