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

Step-by-step explanation:

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

`\text{Moles of} B_2H_6=(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 =
`(3)/(1)* 0.5302=1.591moles` of
`O_2`

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

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