Mg + O2 ------> MgO

Question

if 1.00 g of magnesium is ignited in a flask containing 0.500liters of oxygen at STP, how many grams of magnesium oxide areproduced?
what is the name and the amount of the reactant inexcess?
My work:
1g Mg/24.3g=0.041 mol Mg;
0.5L O2/22.4L=0.022 mol O2; So O2 is the LR; 0.022 molx 56.3g/mol=1.24g MgO2.
0.041mol Mg-0.022mol O2=1.24mol Mg inexcess. Am I right? Thanks!

Answer 1

No. You didn't balance the equation.
2Mg + O2 ==> 2MgO
I'll let you redo it form here.
Thanks for posting your work. It makes it easy to check that way.
Reost here if you want me to recheck.

Answer 2

Thanks a lot!
Here's original question:
If 1.00 of magnesium is ignited in a flask containing 0.500 liter of oxygen at STP, how many grams of magnesium oxide are produced?
My work:
Mg + O2 ------> MgO2, is this right?

Answer 3

0.041mol Mg-0.022mol O2=0.019mol Mg inexcess. Thank you!

Answer 4

Lisa, you have to balance equations with the right compound formulas. 2Mg+O2>>2MgO
moles O2:.5/22.4 =.022moles
moles Mg:Magnesium oxide is MgO 0.041 mol Mg;
but you need twice as many moles of Mg as O2 (see balanced equation), so you need then .0205 moles O2. You have more than that, so the LR is Mg.

how many moles of MgO are produced: same number of moles of Mg: .041

recheck my numbers.

Answer 5

2Mg + O2 ==> 2MgO
1g Mg/24.3g=0.041 molMg, 0.041mol/2mol=0.0205 mol Mg;
0.5L O2/22.4L=0.0223 molO2; So Mg is the LR(0.0205mol Mg<0.0223 molO2);
0.041 molx40.3g MgO/mol=1.65g MgO2.
0.022molO2-0.0205molMg =0.0018 mol O2 inexcess.
Thanks!