June 19, 2024, 01:53:28 PM
Forum Rules: Read This Before Posting

### Topic: Limiting Reagent: Mg + O2  (Read 7009 times)

0 Members and 1 Guest are viewing this topic.

#### zeus

• Guest
##### 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?

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 »

#### Mitch

• General Chemist
• Sr. Member
• Posts: 5298
• Mole Snacks: +376/-3
• Gender:
• "I bring you peace." -Mr. Burns
##### 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 »
Most Common Suggestions I Make on the Forums.
1. Start by writing a balanced chemical equation.
2. Don't confuse thermodynamic stability with chemical reactivity.
3. Forum Supports LaTex