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The number of moles of aluminum that react with 15.7 grams of oxygen according to the equation 4 Al + 3 0 2 - 2 Al 20 3

Sagot :

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]

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