nH2 = 3.36/22.4 = 0.15 (mol)
nFe2O3 = 40/160 = 0.25 (mol)
Fe2O3 + 3H2 -t0-> 2Fe + 3H2O
Bđ: 0.25.......0.15
Pư: 0.05.......0.15.........0.1.......0.15
Kt: 0.2.............0............0.1.......0.15
mCr = mFe2O3(dư) + mFe = 0.2*160 + 0.1*56 = 37.6 (g)
mH2O = 0.15 * 18 = 2.7 (g)