# Magnesium metal stoichiometry?

Magnesium metal (10.21 g) was burned in excess oxygen to

produce magnesium oxide. Determine the initial volume of

oxygen at STP if the volume of oxygen left after the reaction was

6.81 dm3 at STP.

Relevance

Magnesium + Oxygen ➜ Magnesium Oxide

2Mg + O₂ ➜ 2MgO

2 Mg = 2•24.3 = 48.6 g/mol

O₂ = 32

2MgO = 2(16+24.3) = 80.6

48.6 g of Mg react with 32 g of O₂ ➜ 80.6 g of MgO

how much oxygen is required to react with 10.21 g of Mg ?

ratio is 32/48.6

32/48.6 = x/10.21

x = 6.72 g

or 6.72 g / 32g/mol = 0.210 mol

convert that to gas at STP (old standard being used here)

0.210 mol x 22.41 L/mol = 4.71 L

6.81 dm3 = 6.81 L

you started with 6.81 + 4.71 = 11.52 L

One mole of any ideal gas at STP has a volume of 22.41L

(old def of STP, 1 atm)

One mole of any ideal gas at STP has a volume of 22.71L

(new def of STP, 100 kPa)

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• 10.21 / 24.3  =  0.42016 mol

2Mg  +  O2  -->  2MgO

0.42016  X  1/2  X  22.4  =  4.706 L  O2  used in the reaction

4.7016 +  6.81  =  11.52 L  or dm^3  round for sig figs

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