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Mg(ClO3)2-> MgCl2+ 3O2

How many grams of magnesium chlorate would you need to produce 5.00 liters of oxygen?

User Malay
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1 Answer

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The volume of 1 mol of O2 at STP is is 22.4L

Therefore, 5L of O2 at STP is equivalent to (1 mole/22.4 L) x 5 L = 0.2083 mol of O2

From the balanced equation, 3 moles of O2 are produced by heating 1 mole of Mg(ClO3)2

Therefore, 0.2083 mol of O2 is produced by (1/3) x 0.2083 = 0.0694 mol of Mg(ClO3)2.

Molar mass of Mg(ClO3)2 = 191.32 g/mol

Therefore, the mass of Mg(ClO3)2 required is 0.0694 mol x 191.32 g/mol = 13.28 g