Question

Giải câu 31 giùm mình

Answer 1

\(n_O=\dfrac{15.12,8\%}{16}=0,12\left(mol\right)\)

=> n_H2O = 0,12 (mol)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Bảo toàn H: n_HCl = 0,54 (mol)

m_muối = m_{rắn bđ} - m_O + m_Cl = 15 - 0,12.16 + 0,54.35,5 = 32,25 (g)

Answer 2

\(m_{O_2}=12,8\%m_{hh}=12,8\%\cdot15=1,92g\)

\(\Rightarrow n_{O_2}=\dfrac{1,92}{32}=0,06mol\Rightarrow n_O=0,12mol\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15mol\)

BTe: \(2H^++2e\rightarrow H_2\uparrow\)

         0,3                0,15

        \(2H^++O+2e\rightarrow H_2O\)

        0,24    0,12 0,12      0,06

\(n_{H^+}=0,3+0,24=0,54mol\Rightarrow n_{HCl}=0,54mol\)

BTKL: \(m_{hh}+m_{HCl}=m_{muối}+m_{H_2}+m_{H_2O}\)

\(\Rightarrow15+0,54\cdot36,5=m_{muôi}+0,15\cdot2+0,12\cdot18\)

\(\Rightarrow m_{muối}=32,25g\)