How many g of $B_{2} H_{6}$ $B_2H_6$ will react with 3.00 mol of $O_{2}$ $O_2$ ?

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Answer 1 · 2016-10-14T06:36:38

Approx. $28 \cdot g$ $28*g$

We need a stoichiometrically balanced equation:

$B_{2} H_{6} (g) + 3 O_{2} (g) \to 2 B {(O H)}_{3} (s)$ $B_2H_6(g) + 3O_2(g) rarr 2B(OH)_3(s)$

Is this balanced?

And thus we need 1 equiv diborane, i.e. $2 \times 13.83 \cdot g$ $2xx13.83*g$ .

How many g of B2H6B_2H_6 will react with 3.00 mol of O2O_2?