If 12.01 g of Mg metal and 6.56 g of O2 gas are reacted,

Question

1) How many moles of MgO can be produced from the given mass of Mg?
2) How many moles of MgO can be produced from the given mass of O2?
3) From the above calculations, determine which of the reactants is the limiting reagent.
Recall the balanced reaction is:
2 Mg (s) + O2 (g) 2 MgO (s)
Enter the number of moles of MgO from 12.01 of Mg here:

Enter the number of moles of MgO from 6.56 of O2 here:

The limiting reagent is?: Mg O2

Answer 1

Ben,

1.use grams of Mg to moles of MgO by using molar mass of Mg, then mole ratio.
2.same process as number 1 but with O2.
3.determine which element produced less moles of MgO by comparing answer to question 1 and 2; that is your limiting reagent.

which produced less moles of MgO2? that's your answer to which is limiting reagent

Answer 2

see http://www.jiskha.com/display.cgi?id=1459199032