Determine the number of grams H_2 formed when 250.0 mL of 0.743 M HCl solution reacts with 3.41 times 10^23 atoms of Fe according to the following reaction. 2 HCl(aq) + Fe(s) rightarrow H_2(g) + FeCl_2(aq) A) 0.374 g B) 1.33 g C) 1.14 g D) 0.187 g E) 1.51 g
Expert Answer
Answer
We know that
Mol=concentration*volume
Mol of HCl=0.743*0.25=0.186 mol
Now
Mol=no of atoms/avogadro const
So mol of Fe=3.41*10^23/6.022*10^23=0.566 mol
Now according to stoichiometry,
2 mol of HCl=1 mol of Fe
So if we use limiting reactant as Fe than
We required 1.132 mol of HCl.but we have 0.186 mol of HCl.
So limiting reactant is HCl
So HCl mol=0.186
As per reaction
Fe required=0.0929 mol
So Fe in excess
So H2 mol=0.0929 mol
H2 Mass=mol*mol.weight=0.0929*2=0.186 gm