Question

How many grams of if5 would be produced using 44 grams of i2o5 and 101 grams of brf3?

Answer 1

Answer:

Explanation:

I2o5 = 333.81 g/mol  

Brf3 = 136.90 g/mol  

If5 = 221.89 g/mol  

The limiting reagent using conversions will be -

= (44.01 g of I2o5) x (1 mole of I2o5)/(333.81 g) x (20 moles of Brf3)/(6 moles of I2o5)] x (136.90 g Brf3)/(1 mole) = 60.16 g Brf3  

Since, there is more quantity of Brf3 than required, thus I2o5 will be the limiting reagent. Thus the maximum amount of If5 that can be made is -  

= (44.01 g I2o5) x [(1 mole I2o5)/(333.81 g)] x [(12 moles of If5)/(6 moles I2o5)] x [(221.89 g If5)/(1 mole of If5)]

= 58.51 g If5.