1
nH2= 1.68/22.4 =0.075 mol
2K + 2H2O -> 2KOH + H2
0.15 0.075
=> mK = 0.15*39 = 5.85g
=> mFe = mhh - mK= 17.05-5.85 = 11.2g
b/
nH2 p/ứ = 0.075*85% = 0.06375 mol
CuO + H2 -> Cu + H2O
0.06375 0.06375
=>m Cu = 0.06375*64 = 4.08g
3//
nH2 = 4.48/22.4 =0.2 mol
Ba+2H2O -> Ba(OH)2 + H2
0.2 0.2
m Ba = 137*0.2=27.4g
=> mCaO = 33-27.4 =5.6g
b//
nCu = 9.6/6.4=1.5 mol
CuO +H2 - >Cu + H2O
1.5 1.5
=> 2nH2p/ứ = nH = 2*1.5 = 3mol
