Question

When 8.42 g of B₂H₉ reacts with 27.8 g O₂ according to the equation :

  2B₂H₉ + 12O₂  →  5B₂O₃  +  9H₂O

8.24 g H₂O forms. What is the percent yield for the reaction.

Answer 1

Let us consider the reaction, 2 B. На + 1202 5 B2O3 + 9 H₂O 32 32 Molar mass (almol) Amount 27.8. 8.42 mole 0.27 0.87 first we will find the limiting reagent of the reaction. 8.42 mole of B2 2 tq mo) 31 0.27 Mન 27.8 mole of O2 mer 32 0.87 mol 0.27 mole of B2 Hq = I 3,22 Actual ratio mole of oz 0.87 11 Stoichiometric raho 2 mole of B2Hq 12 mole of o2 al ratio , from the stoichiometric B2 Hq I mole of reaut with 6 mote of Oz. available to mole of o2 react with B₂ Hq . Here only 3.22 a lie it is completely So, O2 i's the limiting reagent reaution in the congumed

* Then theoritical yeild of H₂O = 0.87 0.87 mol of O2 9 mol of H₂O x 12 mol of O2 = 0.65 25 mol of H20 0065 25 X 18 १ of H₂O Molar mass of W H20 = 18 g/mor) = l lo 7 4 5 g of H₂O

Now per cent yeild actual yeild of H2O theoritical yeird of H29 8:24 11.745 X 100% - 70.15 % M = 70015% of so reaction percent geild