Question

Consider the reaction

$B_2{H}_6\left(g\right)+3O_2\left(g\right)\to B_2{O}_3\left(s\right)+3H_{2}O\left(g\right)\Delta H=-2035kJ$

Calculate the amount of heat released when 54.0 g of diborane is combusted.

Short answer

The heat released from 55 g of B₂H₆ is 4046.5 kJ.

Step-by-step solution

Heat Released from 1 mole

Here heat released from 1 mole of B₂H₆ i.e., 27.66 g is given 2035 kJ., here the negative sign shows that the energy is being released.

$\begin{array}{l}{B}_2{H}_6\left(g\right)+3O_2\left(g\right)\to B_2{O}_3\left(s\right)+3H_{2}O\left(g\right)\Delta H=-2035kJ\\ 1mole\\ 27.66g\end{array}$

Heat Released from 54 g Borane

$\begin{array}{l}Heatreleasedfrom1mole(27.66g)diborane=-2035kJ\\ Heatreleasedfrom55gdiboranewillbe=\frac{-2035}{27.66}\times 55kJ=4046.5kJ\end{array}$

Hence, the heat released from 55 g of B₂H₆ is 4046.5 kJ.