A metal form two oxides upon reaction with oxygen. In first compound 8 g of oxygen reacts with 29 g of metal. In second compound, 3 g of oxygen reacts with 7.25 g of metal. If the formula for the first compound is MO, then what is the formula for second?

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Answer 1 · 2018-07-03T12:11:50

We have to find the amount of oxygen that would react with $29g$ of the metal in the second compound.

$29div7.25=4$

So for $29g$ of the metal in the second compound we would need:
$4xx3g=12g$ of oxygen.

So the second oxide has $12gdiv8g=1.5$ times as much oxigen:

Formula: $MO_(1.5)$ which will be $M_2O_3$