Đáp án:
\(\text C\%_{HCl}=4,867\%;\rm{C\%_{dd A}}=8,17\%\)
Giải thích các bước giải:
\(Fe+2HCl\to FeCl_2+H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3\ \rm{mol}\to n_{HCl}=0,3\cdot 2=0,6\ \rm{mol}\to m_{\text{HCl}}=0,6\cdot 36,5=21,9\ \rm{gam}\)
\(\to \text C\%_{\text{dd HCl}}=\dfrac{21,9}{450}\cdot 100\%=4,867\%\)
\(n_{Fe}=n_{H_2}=0,3\ \rm{mol}\)
\(\to m_{Fe}=0,3\cdot 56=16,8\ \rm{gam}\)
BTKL: \(m_A=16,8+450-0,3\cdot 2=466,2\ \rm{gam}\)
\(n_{FeCl_2}=n_{H_2}=0,3\ \rm{mol}\)
\(\to m_{FeCl_2}=127\cdot 0,3=38,1\ \rm{gam}\)
\(\to \text C\%_{\text{dd A}}=\dfrac{38,1}{466,2}\cdot 100\%=8,17\%\)
