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Answer 1 · 2017-03-01T17:23:02

$1.19090412854 gram of O_{2}$ $"1.19090412854 gram of" O_2$

1 mole of MgO has 1 mole of Mg and 1 mole $O_{2}$ $O_2$

amount of Mg in 3 grams of MgO

Molar mass of MgO = 24.305mol + 15.999mol = 40.304mol

Moles in 3grams = $\frac{3g}{40.304mol/gram} = 0.07443429932 m o l$ $"3g"/"40.304mol/gram" = 0.07443429932mol$

Mole Ratio of O to MgO

1mol : 1mol

Therefore amount of oxygen in MgO

0.07443429932mol : 0.07443429932mol

Therefore there is 0.07443429932 moles of $O_{2}$ $O_2$

Weight of 0.07443429932moles of $O_{2}$ $O_2$ =

$0.07443429932moles \cdot 15.9994mol/gram = 1.19090412854 gram of O_{2}$ $"0.07443429932moles" * "15.9994mol/gram" = "1.19090412854 gram of" O_2$