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.

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kobenhavn kobenhavn · Answer 1 · 2020-04-19T17:28:09+02:00

Answer: 50.91 grams

Explanation:

[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]

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

The balanced chemical equation is :

[tex]B_2H_6(g)+3O_2(g)\rightarrow B_2O_3(s)+3H_2O(l)[/tex]

According to stoichiometry :

1 mole of [tex]B_2H_6[/tex] require = 3 moles of [tex]O_2[/tex]

Thus 0.5302 moles of [tex]B_2H_6[/tex] will require =[tex]\frac{3}{1}\times 0.5302=1.591moles[/tex] of [tex]O_2[/tex]

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

Thus 50.91 grams of [tex]O_2[/tex] reacts with 14.67 g of diborane