I_2 O_5 + BrF_3 rightarrow IF_5 + O_2 + Br_2 What is the balanced equation expre

Question

I_2 O_5 + BrF_3 rightarrow IF_5 + O_2 + Br_2 What is the balanced equation expressed using the smallest whole number coefficients? A chemist needs to make 120.0 g of IF_5. If the chemist reacts 90.26 g of I_2O_5 and 202.0 g of BrF_3, will the chemist make his quota? Some molar masses I_2 O_5 = 333.81 g/mol; BrF_3 = 136.9 g/mol; IF_5 = 221.89 g/mol What reactant is in excess, and how many g of it remain at the end of the reaction? Show your work. Suppose that instead of going to completion, the reaction only produces 85% yield. How much IF_5 is formed (using the same initial reactant amounts as in part b)? Did the chemist meet his quota under these conditions?

Explanation / Answer

1) Balanced equation : 6I2O5 + 20BrF3 ---> 2IF5 + 15O2 + 10Br2

2) moles of IF5 to be formed = 120/221.89 = 0.54 mols

moles of I2O5 taken = 90.26/333.81 = 0.27 mols

moles of BrF3 taken = 202/136.9 = 1.47 mols

limiting reagent

For complete consumption of I2O5 we would require = 0.27 x 20/6 = 0.9 mols of BrF3

For complete consumption of BrF3 we would need = 1.47 x 6/20 = 0.44 mols of I2O5

Since mols of I2O5 available is less then required, this is the limiting reagent

mass of IF5 formed = 0.27 x 12 x 221.89/6 = 120 g

So it will give the desired amount of IF5

3) BrF3 is in excess

mass of BrF3 remaining = (1.47 - 0.9) x 136.9 = 78.03 g

4) If yield of reaction is 85%

mass of IF5 formed = 0.85 x 120 = 102 g

No the chemist would not met his required criteria of 120 g IF5 in this case

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