Ba+2H2O\(\rightarrow\)Ba(OH)2+H2 (1)
BaO+H2O\(\rightarrow\)Ba(OH)2 (2)
nH2=\(\frac{\text{1,912}}{22,4}\)=0,085 mol
\(\rightarrow\) nBa=0,085 mol; nBa(OH)2 (1)= 0,085 mol
nCO2=\(\frac{\text{6,272}}{22,4}\)=0,28 mol
Ba(OH)2+CO2\(\rightarrow\)BaCO3+H2O
\(\rightarrow\) nBa(OH)2=0,28 mol
nBa(OH)2 (1)=0,085 mol \(\rightarrow\) \(\text{nBa(OH)2=0,195 mol= nBaO}\)
\(\rightarrow\) \(\text{a=mBa+mBaO=0,085.137+0,195.153= 41,48g}\)