looks good to me, I didn't do the calculator punching.
I would personally have balanced the equation with whole numbers, but it looks ok.
Mg + O2 ------> MgO
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 mol O2x56.3g MgO2=1.24 g Mgo2;
0.041molMg-0.022molO2=0.019 mol Mg or 0.462g inexcess.
3 answers
See my response to your duplicate post above. Since you didn't balance the equation I think all of your work needs to be redone. For openers I don't believe O2 is the LR.
if 1.2.5 g of magnesium is ignited in a flask containg 2.500 liters of oxygen at STP which is the limiting reactant
what is the theoretical yield of Magnesium oxide
if the % yield is only 72.5% what is the actual yield of Magnesium oxide
what is the theoretical yield of Magnesium oxide
if the % yield is only 72.5% what is the actual yield of Magnesium oxide