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Chemistry 18 Online
OpenStudy (anonymous):

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?

OpenStudy (wolfe8):

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/

OpenStudy (anonymous):

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 :)

OpenStudy (joannablackwelder):

Looks great so far. Can you use stoichiometry on both of those numbers of moles to determine which one would produce less MgO?

OpenStudy (joannablackwelder):

The one that produces less of the product will be "used up" first in the reaction and would be the limiting reactant.

OpenStudy (anonymous):

For magnesium: 4.66 g MgO For oxygen: 5.59 g Mg) So magnesium would be the limiting reactant, and that answers A. :D

OpenStudy (anonymous):

MgO, oops

OpenStudy (joannablackwelder):

Awesome! :) You can get c directly...

OpenStudy (anonymous):

Great! I'm a bit confused for what to do for B, though :P

OpenStudy (joannablackwelder):

Ok, that one takes a bit more work. Can you take the amount of MgO produced and back-calculate the amount of O2 consumed?

OpenStudy (joannablackwelder):

In excess means left over.

OpenStudy (joannablackwelder):

That is MgO produced from the limiting reactant.

OpenStudy (anonymous):

1.85 g of O2 were consumed. :3

OpenStudy (joannablackwelder):

Great! So, if you started with 2.22g and 1.85 were consumed, how many do you have left over?

OpenStudy (anonymous):

0.37 g :)

OpenStudy (joannablackwelder):

:)

OpenStudy (anonymous):

I'll try those answers again, since I've gotten the same answers, and my homework program says I'm wrong :P

OpenStudy (joannablackwelder):

Hmm, are you watching sig figs?

OpenStudy (joannablackwelder):

Looks like the problem has 4 sig figs.

OpenStudy (anonymous):

Ah, now it went through :D Thanks so much! :)

OpenStudy (joannablackwelder):

No worries. Keep up the good work!

OpenStudy (wolfe8):

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 :)

OpenStudy (anonymous):

Thanks, wolfe! :)

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