Answer:
Explanation:
1mol of [tex]O_2=2*16{gr\over{mol}}=32{gr\over{mol}}\\\\15.7gr->15.7gr*{1\over{32{gr\over{mol}}}}=0.491mol~of~O_2[/tex]
as 1mol of molecular oxigen reacts with 4 mol of aluminium
1 mol of O2 -----------------------------> 4 mol of Al
0.491 mol of O2 ------------------------------> x
[tex]x={0.491*4\over{1}}~mol~of~Al=1.9625~mol~of~Al[/tex]
