Calculate the mass (g) of each product formed when 43.82 $\mathrm{g}$ of diborane $\left(\mathrm{B}_{2} \mathrm{H}_{6}\right)$ reacts with excess water:

$$

\mathrm{B}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{BO}_{3}(s)+\mathrm{H}_{2}(g)[\text { unbalanced }]

$$

a)

$\mathrm{m}=\mathrm{n}\left(\mathrm{H}_{3} \mathrm{BO}_{3}\right)^{*} \mathrm{Mr}\left(\mathrm{H}_{3} \mathrm{BO}_{3}\right)=195,94 \mathrm{g}$

b)

$\mathrm{m}\left(\mathrm{H}_{2}\right)=\mathrm{n}^{*} \mathrm{Mr}\left(\mathrm{H}_{2}\right)=9,52^{*} 2=19,04 \mathrm{g}$

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University of Maryland - University College

University of Kentucky

Numerade Educator

Brown University

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