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3 years ago

From the unbalanced reaction: B2H6 + O2 ---> HBO2 + H2O

Chemistry

1 answer:

Drupady [299]3 years ago

4 0

Answer: 125 g

Explanation:

To calculate the moles :

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

$\text{Moles of} B_2H_6=\frac{36.1g}{17}=1.30moles$

The balanced reaction is:

$B_2H_6+3O_2\rightarrow 2HBO_2+2H_2O$

According to stoichiometry :

1 mole of $B_2H_6$ require = 3 moles of $O_2$

Thus 1.30 moles of $B_2H_6$ will require= $\frac{3}{1}\times 1.30=3.90moles$ of $O_2$

Mass of $O_2=moles\times {\text {Molar mass}}=3.90moles\times 32g/mol=125g$

Thus 125 g of $O_2$ will be needed to burn 36.1 g of $B_2H_6$

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Now put all the given values in the above expression, we get

$\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]$

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