Câu hỏi của 🍀 Bé Bin 🍀

\(PTHH:Fe+2HCl->FeCl_2+H_2\uparrow\) 0,2->0,4--------->0,2----->0,2 (mol)\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)\(n_{FeCl_2}=0,2\cdot1:1=0,2\left(mol\right)\)\(m_{FeCl_2}=n\cdot M=0,2\cdot\left(56+35,5\right)=18,3\left(g\right)\)\(n_{HCl}=0,2\cdot2:1=0,4\left(mol\right)\\ m_{HCl}=n\cdot M=0,4\cdot\left(1+35,5\right)=14,6\left(g\right)\)Câu trả lời: \(PTHH:Fe+2HCl->FeCl_2+H_2\uparrow\) 0,2->0,4--------->0,2----->0,2 (mol)\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)\(n_{FeCl_2}=0,2\cdot1:1=0,2\left(mol\right)\)\(m_{FeCl_2}=n\cdot M=0,2\cdot\left(56+35,5\right)=18,3\left(g\right)\)\(n_{HCl}=0,2\cdot2:1=0,4\left(mol\right)\\ m_{HCl}=n\cdot M=0,4\cdot\left(1+35,5\right)=14,6\left(g\right)\)