For the single replacement reaction between lithium iodide and oxygen?

how many moles of your excess reactant remain at the end of the reaction?

Update:

if you start with 54g of lithium iodide and 5.0 g of oxygen, how many moles of reactant remain at your solution

1 Answer

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  • 8 months ago

    Balanced equation:

    4 LiI + O2 --> 2 Li2O + 2 I2

    Moles LiI = 54 g / 133.85 g/mol = 0.403 mol LiI

    Moles O2 = 5.0 g / 32.0 g/mol = 0.156 mol O2

    0.403 mol LiI X (1 mol O2/ 4 mol LiI) = 0.101 mol O2

    Because you begin with more O2 than this, LiI is the limiting reactant, and 0.156 - 0.101 = 0.055 mol of O2 remains when the reaction is complete. 

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