# If a 0.050g sample of the boron hydride burns completely in O_2, what will be the pressure of the gaseous water In a 4.25-L flask at 30.0°C?

Dec 10, 2015

I assume that you mean diborane, ${B}_{2} {H}_{6}$.
${B}_{2} {H}_{6} \left(g\right) + 3 {O}_{2} \rightarrow {B}_{2} {O}_{3} \left(s\right) + 3 {H}_{2} O \left(g\right) \uparrow$

#### Explanation:

You are presumed to have (0.050*g)/(27.67*g*mol^-1) = ?? mol diborane.

By the stoichiometry of the reaction listed, for each mole of diborane that combusts, $3 \cdot m o l$ of water are produced.

Assuming ideality, $P$ $=$ $\frac{n R T}{V}$ $=$ $3 \times \left(\frac{0.050 \cdot g}{27.67 \cdot g \cdot m o {l}^{-} 1}\right) \left(\times 0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 m o {l}^{-} 1 \times 303 K\right) \times \left(\frac{1}{4.25 \cdot L}\right)$ $=$ ?? atm.

Why did I use $303 \cdot K$ instead of the given temperature in Centigrade? Would it make a difference if I used the formula $B {H}_{3}$ instead of that of diborane?

Please note that this answer rests on the assumption that you mean diborane. There are many other borohydrides.