Answer:
[tex]m_{HBO_2}=441.8gHBO_2[/tex]
Explanation:
Hello there!
In this case, since the combustion of B2H6 is:
[tex]B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O[/tex]
Thus, since there is 1:2 mole ratio between the reactant and product, the produced grams of the latter is:
[tex]m_{HBO_2}=139.5gB_2H_6*\frac{1molB_2H_6}{27.67gB_2H_6} *\frac{2molHBO_2}{1molB_2H_6} *\frac{43.82gHBO_2}{1molHBO_2}[/tex]
[tex]m_{HBO_2}=441.8gHBO_2[/tex]
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