You are performing a reaction with 1.7 moles of hydroiodic acid and 3.43 moles of zinc bromide: 2hi + znbr2 → 2hbr + zni2. how many moles of zinc iodide can be made?

number of moles of HI = 1.7 mol

number of moles of ZnBr2 = 3.43 mol

2HI + ZnBr2 → 2HBr + ZnI2

every 2 mole of HI will require 1 mole of ZnBr2

mole of ZnBr2 that will react = 1.7 / 2 = 0.85 mol

Therefore HI is the limiting reactant and not all ZnBr2 will be reacted

For every 2 mole of HI will form one mole ZnI2

1.7/2 = 0.85 mol of ZnI2 formed

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