Question

A scientist studying the reaction between decaborane and oxygen mixed 65.0 g of B₁₀H₁₈ with 165.0 g of O₂. This reaction generates B₂O₃ and H₂O as the only products. Compute how many grams of B₂O₃ are present after the reaction went to completion.

Answer 1

number of mole = (given mass)/(molar mass)
molar mass of B10H18 = 126.3 g/mol
molar mass of O2 = 32 g/mol
number of mole of B10H18 = 65/126.3
= 0.52 mole
number oof mole of O2 = 165/32
= 5.16 mole

reaction taking place is
B10H18 + 12O2 -- > 5B2O3 + 9H2O
according to reaction
1 mole of B10H18 required 12 mole of O2
0.52 mole of B10H18 required (12*0.52) mole of O2
0.52 mole of B10H18 required 6.24 mole of O2
but we have only 5.16 mole of O2
so, O2 is limiting reagent
also,
12 mole of O2 give 5 mole of B2O3
1 mole of O2 give 5/12 mole of B2O3
5.16 mole of O2 give (5/12)*5.16 mole of B2O3
5.16 mole of O2 give 2.15 mole of B2O3
number of mole of B2O3 produce = 2.15 mole
molar mass of B2O3 = 69.6 g/mol
mass of B2O3 produce = (number of mole of B2O3)*(molar mass of B23)
= 2.15*69.6
= 149.6 g

Answer : 149.6 g