PTHH: \(Fe+2HCl-->FeCl_2+H_2\uparrow\)
\(n_{HCl}=2.0,2=0,4mol=>n_{Fe}=0,2mol;n_{FeCl_2=0,2mol}\)
a) mFe-phản-ứng:0,2.56=11,2gam
b) mmuối-khan:0,2.127=25,4gam
c) PTHH:\(2HCl+Ba\left(OH\right)_2-->BaCl_2+2H_2O\)
\(n_{HCl}=2.0,1=0,2mol\)
theo PTHH=> \(n_{Ba\left(OH\right)_2}=0,1mol\)
=> \(V_{Ba\left(OH\right)_2}=\dfrac{0,1}{1}=0,1lit=100ml\)