Question

Phân huỷ toàn Fe(OH)3 thu đc 16(g) Fe₂O₃ và b(g) H₂O. Tính a,b

Answer 1

\(n_{Fe_2O_3}=\dfrac{16}{160}=0.1\left(mol\right)\)

\(2Fe\left(OH\right)_3\underrightarrow{^{^{t^0}}}Fe_2O_3+3H_2O\)

\(0.2...................0.1..........0.3\)

\(m_{Fe\left(OH\right)_3}=a=0.2\cdot107=21.4\left(g\right)\)

\(m_{H_2O}=b=0.3\cdot18=5.4\left(g\right)\)

Answer 2

PTHH:      \(2Fe\left(OH\right)_3\underrightarrow{t^o}Fe_2O_3+3H_2O\)

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)

\(n_{Fe_2O_3}=n_{H_2O}=0,1\left(mol\right)\)

\(m_{H_2O}=0,1.18=1,8\left(g\right)\)

\(b=1,8\)