Chemistry help

Question

The compound diborane (B2H6) was at one
time considered for use as a rocket fuel. Its
combustion reaction is
B2H6(g) + 3 O2(l) → 2 HBO2(g) + 2 H2O(l)
The fact that HBO2, a reactive compound,
was produced rather than the relatively inert
B2O3 was a factor in the discontinuation of
the investigation of the diborane as a fuel.
007 (part 1 of 2)What mass of liquid oxygen (LOX) would be
needed to burn 226.3 g of B2H6?
Answer in units of g.

008 (part 2 of 2) 10.0 points
What mass of HBO2 is produced from the
combustion of 61.3 g of B2H6?
Answer in units of g.

Answer 1

This is a stoichiometry question, so always start with the balanced equation and write what you know and what you want to get underneath.

 

1 B₂H₆(g) + 3 O₂(l) → 2 HBO₂(g) + 2 H₂O(l) <-- This is the equation you gave, so I assume it is correctly balanced.

 

226.3g        ?g

 

You need to go from the mass of one chemical to the mass of another chemical, so you will need to go through moles in order to "switch" from one chemical to the other. Remember that coefficient ratios in an equation are really mole to mole ratios.

 

So, the plan is to go g B₂H₆ -(1)-> mol B₂H₆ -(2)-> mole O₂ -(3)-> g O₂

 

?g O₂ = 226.3g B₂H₆ x (1mol B₂H₆/27.67g B₂H₆) x (3 mol O₂/1 mol B₂H₆) x (32.00 g O₂/mol O₂) = _____g O₂

 

         = (226.3*3*32/27.67) g O₂

 

------------------------------------------------------------------

 

The process is the same in the second part of the question

 

1 B₂H₆(g) + 3 O₂(l) → 2 HBO₂(g) + 2 H₂O(l) 

61.3g                         ?g

 

This time the plan will be g B₂H₆ -(1)-> mol B₂H₆ -(2)-> mole HBO₂ -(3)-> g HBO₂

 

?g HBO2 = 63.1gB₂H₆ x (1mol B₂H₆/27.67g B₂H₆) x (2 mol HBO₂/1 mol B₂H₆) x (43.82 g HBO₂/mol HBO₂)

 

 

              = (63.1*2*43.82/27.67) gHBO₂

 

I hope this helps. steps are color coded and the numbers in bold are the coefficients in the chemical equation.