2B5H9+ 12O2-----> 5B2O3+ 9H20
How many moles of B2O3 can theoretically be obtained from 4 moles of B5H9 and excess O2?
How much B5H9 must be reacted with O2 to produce exactly 100 g of B2O3?
120 g of B5H9 and 200g of 02 are reacted. How much B2O3 will be produced?
150g of B2O3 were obtained. WHat is the percent yield in this reaction?