a)CH3OH + 3/2 O2 ---> CO2 + 2 H2O
b)puritatea=masa pura/masa impura*100=>masa pura=puritatea*masa impura/100=>masa pura = 200*64/100=128 g CH3OH pur
32 g CH3OH.........3/2*32 g O2
128 g CH3OH.............x g O2 x = 192 g O2
nr moli O2=masa O2/masa molara O2=192/32=6 moli O2
pV=nr moli*R*T=>V=nr moli*R*T/p=6*0,082*300/1=>V oxigen = 147,6 litri
V aer = 5 * V oxigen = 5*147,6=>V aer = 738 litri