Question

Answer 1

Ta có: \(n_{HCl}=0,2.1,5=0,3\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

__0,15___0,3_____0,15___0,15 (mol)

a, m_Fe = 0,15.56 = 8,4 (g)

b, V_H2 = 0,15.22,4 = 3,36 (l)

c, Ta có: \(n_{Fe_2O_3}=\dfrac{64}{160}=0,4\left(mol\right)\)

PT: \(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

Xét tỉ lệ: \(\dfrac{0,4}{1}>\dfrac{0,15}{3}\), ta được Fe₂O₃ dư.

Theo PT: \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{2}{3}n_{H_2}=0,1\left(mol\right)\\n_{Fe_2O_3\left(pư\right)}=\dfrac{1}{3}n_{H_2}=0,05\left(mol\right)\end{matrix}\right.\)

⇒ n_{Fe2O3 (dư)} = 0,35 (mol)

⇒ m _{chất rắn} = m_Fe + m_{Fe2O3 (dư)} = 0,1.56 + 0,35.160 = 61,6 (g)

Bạn tham khảo nhé!