CTHH: FexOy
Gọi số mol CO2 sinh ra là x (mol)
\(n_{NaOH}=1.1=1\left(mol\right)\)
\(m_{dd.NaOH}=1,0262.1000=1026,2\left(g\right)\)
=> mdd sau pư = 1026,2 + 44x (g)
PTHH: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)
1------>0,5------->0,5
\(Na_2CO_3+CO_2+H_2O\rightarrow2NaHCO_3\)
(x-0,5)<---(x-0,5)----------->(2x-1)
=> \(\left\{{}\begin{matrix}n_{Na_2CO_3}=1-x\left(mol\right)\\n_{NaHCO_3}=2x-1\left(mol\right)\end{matrix}\right.\)
Ta có: \(\dfrac{\left(1-x\right).106+\left(2x-1\right).84}{1026,2+44x}.100\%=6,47\%\)
=> x = 0,75 (mol)
PTHH: \(Fe_xO_y+yCO\underrightarrow{t^o}xFe+yCO_2\)
\(\dfrac{0,75}{y}\)<----------------0,75
\(Fe_xO_y+yH_2SO_4\rightarrow Fe_x\left(SO_4\right)_y+yH_2O\)
\(\dfrac{0,75}{y}\)---------------->\(\dfrac{0,75}{y}\)
=> \(\dfrac{0,75}{y}\left(56x+96y\right)=103,5\)
=> x : y = 3 : 4
=> CTHH: Fe3O4
