For the single replacement reaction between lithium iodide and oxygen?
how many moles of your excess reactant remain at the end of the reaction?
if you start with 54g of lithium iodide and 5.0 g of oxygen, how many moles of reactant remain at your solution
- hcbiochemLv 72 months ago
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.