In an experiment 2.810 g of magnesium is reacted with 2.220 g of oxygen gas according to the equation: Mg + O2 ==> MgO. A. What is the limiting reactant? B. What mass of the other reactant is in excess? C. What mass of MgO is produced?
Welcome to OpenStudy. First, make sure the equation is balanced. Next, can you find the number of moles of whatever substance you have? Use this: http://www.ptable.com/
The balanced equation is 2Mg + O2 -> 2MgO The number of moles of magnesium is 0.1156 The number moles of oxygen is 0.0694 And thank you :)
Looks great so far. Can you use stoichiometry on both of those numbers of moles to determine which one would produce less MgO?
The one that produces less of the product will be "used up" first in the reaction and would be the limiting reactant.
For magnesium: 4.66 g MgO For oxygen: 5.59 g Mg) So magnesium would be the limiting reactant, and that answers A. :D
MgO, oops
Awesome! :) You can get c directly...
Great! I'm a bit confused for what to do for B, though :P
Ok, that one takes a bit more work. Can you take the amount of MgO produced and back-calculate the amount of O2 consumed?
In excess means left over.
That is MgO produced from the limiting reactant.
1.85 g of O2 were consumed. :3
Great! So, if you started with 2.22g and 1.85 were consumed, how many do you have left over?
0.37 g :)
:)
I'll try those answers again, since I've gotten the same answers, and my homework program says I'm wrong :P
Hmm, are you watching sig figs?
Looks like the problem has 4 sig figs.
Ah, now it went through :D Thanks so much! :)
No worries. Keep up the good work!
Sorry my housemate needed urgent help with something. Good job you two :) Welcome again to OS Anita. Do take time to read our CoC: http://openstudy.com/code-of-conduct and enjoy your stay. If you are satisfied with any question, please reward a medal to whoever you think deserves it and close the question. We also have a fan and testimony system. See you around :)
Thanks, wolfe! :)
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