Trả lời:
\(n_{FeO}=\dfrac{10,8}{72}=0,15\left(mol\right)\)
\(m_{HCl}=\dfrac{150.18,25}{100}=27,375\left(g\right)=>n_{HCl}=\dfrac{27,375}{36,5}=0,75\left(mol\right)\)
PTHH :
\(FeO+2HCl-->FeCl_2+H_2O\)
\(\dfrac{0,15}{1}< \dfrac{0,75}{2}\)
\(=>FeO\) hết , \(HCl\) dư , tính theo FeO
a,
\(n_{FeCl_2}=n_{FeO}=0,15\left(mol\right)=>m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
b, Sau phản ứng HCl dư nên :
\(m_{HCl_{dư}}=m_{HCl_{bđ}}-m_{HCl_{pư}}=36,5.0,75-36,5.0,3=16,425\left(g\right)\)
c, H\(_2\)SO\(_4\) phản ứng với FeO
\(FeO+H_2SO_4-->FeSO_4+H_2O\)
\(n_{H_2SO_4}=n_{FeO}=0,15mol=>m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)
\(=>m_{ddH_2SO_4}=\dfrac{14,7.100}{20}=73,5\left(g\right)\)
