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Topic: Limiting Reagent: Mg + O2  (Read 5551 times)

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zeus

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Limiting Reagent: Mg + O2
« on: April 23, 2004, 12:06:02 PM »
 ???

Which element is in excess when 3.23 x 10^-8g of magnesium is ignited in 2.20 x 10^-9g of pure oxygen? What mass of magnesium oxide is formed?

My answer:

2Mg + O2--->2MgO

Magnesium
3.23 x 10 ^-8g Mg x 1molMg/24.31Mg x 2molMfO/2molMg x 40.31MgO/1molMgO = 2.14 x 10
(3.23 x 10^-8)(2)(40.31)/(24.31)(2) = 2.60 x 10^11/48.62 = 5.34 x 10^10


Oxygen
2.20 x 10^9 x 1molO/32gO x 2molMgO/1molO2 x 40.31MgO/1molMgO = 5.54 x 10^10
(2.20 x 10^9)(2)(40.31)/(32)(1) = 1.77 x 10^12/32 = 5.53 x 10^11

Can someone let me know if these are correct.  ???
« Last Edit: April 24, 2004, 06:20:26 PM by hmx9123 »

Offline Mitch

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Re: Limiting Reagent: Mg + O2
« Reply #1 on: April 23, 2004, 01:31:55 PM »
the first formula for Magnesium looks okay. Oxygen is O2 and you have to use the MW of O2.
« Last Edit: April 24, 2004, 06:20:54 PM by hmx9123 »
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