Ta có:\(\left\{{}\begin{matrix}n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\\n_{H_2SO_4}=\dfrac{100.12,25\%}{100\%.98}=0,125\left(mol\right)\end{matrix}\right.\)
\(PT:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
cứ:: 1..............1.................1.................1 (mol)
Vậy: 0,1---------->0,1------->0,1-------->0,1(mol)
=> mH2=n.M=0,1.2=0,2(g)
Chất dư là H2SO4
Số mol H2SO4 dư là: 0,125-0,1=0,025(mol)
b) Chất sau phản ứng gồm: \(\left\{{}\begin{matrix}FeSO_4\left(0,1mol\right)\\H_2SO_{4\left(dư\right)}\left(0,025mol\right)\end{matrix}\right.\)
=> mFeSO4=n.M=0,1.152=15,2(g)
mH2SO4(dư)=n.M=0,025.98=2,45(g)
mdung dịch sau phản ứng=mFe + md d H2SO4-mH2=5,6+100-0,2=105,4(g)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{m_{FeSO_4}.100\%}{m_{ddsauphanung}}=\dfrac{15,2.100}{105,4}\approx14,42\left(\%\right)\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{m_{H2SO4\left(dư\right)}.100\%}{m_{ddsauphanung}}=\dfrac{2,45.100}{105,4}\approx2,32\left(\%\right)\)
