n Mg=0,2(mol)
2Mg+O2---.2MgO
0,2---0,1----0,2(mol)
a) m MgO=0,2.40=8(g)
b) V O2=0,1.22,4=2,24(l)
Vkk=2,24.5=11,2(l)
PTHH: 2Mg + O2 \(\overset{t^O}{--->}\) 2MgO
a, nMg = \(\frac{4,8}{24}\) = 0,2 mol
Theo PTHH có : nMgO = nMg = 0,2 mol
=> mMgO = 0,2 . 40 = 8 g
b, Theo PTHH có : nO2 = \(\frac{1}{2}\) nMg = 0,1 mol
=> VO2 = 2,24 lit
Mà \(\frac{V_{O2}}{V_{KK}}=\frac{1}{5}\) => VKK = 2,24 .5 = 11,2 lit