\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
\(n_{H_2SO_4\left(bđ\right)}=\dfrac{19,6\%.200}{98}=0,4\left(mol\right)\)
\(n_{Fe}=n_{H_2SO_4\left(pứ\right)}=0,2\left(mol\right)\)
=> \(n_{H_2SO_4\left(dư\right)}=0,4-0,2=0,2\left(mol\right)\)
\(m_{ddsaupu}=11,2+200-0,2.2=210,8\left(g\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{0,2.98}{210,8}.100=9,3\%\)
PTHH: Fe+H2SO4→H2+FeSO4
nFe=\(\dfrac{11,2}{56}\)=0,2(mol)
mH2SO4=\(\dfrac{19,6\%.200}{100\%}\)=39,2(g)
nH2SO4=\(\dfrac{39,2}{98}\)=0,4(mol)
Vì Fe đã pứng hết,theo PTHH ta có:nFe=nH2SO4=0,2(mol)
⇒nH2SO4 dư=0,4-0,2=0,2(mol)
⇒mH2SO4 dư=0,2.98=19,6(g)
m dung dịch sau pứng:200+11.2=211,2(g)
⇒C%H2SO4 sau pứng=\(\dfrac{19,6.100\%}{211,2}\)(xấp xỉ) 9,3%
