Question

Diborane (B2H6) is a gas at room temperature that forms explosive mixtures with air. It reacts with oxygen according to the following equation (which may or may not be balanced): _____ B2H6 (g) + _____ O2 (g) → _____ B2O3 (s) + _____ H2O (l) How many grams of O2 (molar mass 32.00 g/mol) will react with 14.67 grams of diborane (molar mass 27.67 g/mol). Your answer must be expressed to the correct number of significant figures, and with the correct unit.

Answer 1

Answer: 50.91 grams

Explanation:

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$    

$\text{Moles of} B_2H_6=\frac{14.67g}{27.67g/mol}=0.5302moles$

The balanced chemical equation is :

$B_2H_6(g)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)$

According to stoichiometry :

1 mole of $B_2H_6$ require  = 3 moles of $O_2$

Thus 0.5302 moles of $B_2H_6$ will require =$\frac{3}{1}\times 0.5302=1.591moles$ of $O_2$

Mass of $O_2=moles\times {\text {Molar mass}}=1.591moles\times 32.00g/mol=50.91g$

Thus 50.91 grams of $O_2$ reacts with 14.67 g of diborane