n_{Mg}={15}/{24}=0,625(mol)

2Mg+O_2  $$\xrightarrow{t^o}$$  2MgO

0,625->0,3125->0,625(mol)

a) m_{MgO}=0,625.40=25(g)

b) V_{kk}={0,3125.22,4}/{20\%}=35(l)

a)

n_{Mg}=(15)/(24)=0,625(mol)

$$2Mg+O_2\xrightarrow[]{t^o}2MgO$$

$$\xrightarrow[]{TheoPTHH}$$n_{MgO}=1.0,625=0,625(mol)

=>m_{MgO}=0,625.40=25(g)

b)

$$\xrightarrow[]{TheoPTHH}$$n_{O_2}=1/(2).0,625=0,3125(mol)

->V_{O_2}=0,3125.22,4=7(l)

=>V_{kk}=7.5=35(l)

       #Bear