nH2=0,3(mol)
Ta có:
nCl=nHCl=2nH2=0,6(mol)
mCl=35,5.0,6=21,3(g)
mmuối=mKL + mCl=17,7+21,3=39(g)
Vdd HCl=\(\dfrac{0,6}{0,1}=6\left(lít\right)\)
nH2=0,3(mol)
Ta có:
nCl=nHCl=2nH2=0,6(mol)
mCl=35,5.0,6=21,3(g)
mmuối=mKL + mCl=17,7+21,3=39(g)
Vdd HCl=