Question

Starting with the following equation, BCl₃(g) + LiAlH₄(s) → B₂H₆(g) + LiAlCl₄(s) calculate the moles of BCl₃ that will be required to produce 365 grams of B₂H₆.

Answer 1

First balance the equation, then calculate how many moles of B2H6 you need, so n=m/M. Then you can make a proportion: moles of B2H6 in the equation —moles you need to make; moles of BCl3 in the equation — x moles. so then u take n of how much of B2H6 you need to make, multiply with n of BCl3 divided by moles of B2H6 in the equation and you get the answer. I hope that helps :)

Answer 2

26.39 moles of BCl₃ will be required to produce 365 grams of B₂H₆.

What is a mole ?

A mole is defined as 6.02214076 × 10²³ atoms, molecules, ions, or other chemical units.

To determine the moles of BCl₃ required we have to first balance the equation given : BCl₃(g) + LiAlH₄(s) → B₂H₆(g) + LiAlCl₄(s)

The balanced equation will be 4BCl₃ +3 LiAlH₄ → 2B₂H₆ + 3LiAlCl₄

So for each 2 moles of B₂H₆ you need 4 moles of BCl₃ .

no. of moles = (mass given)/(molecular mass)

Given is 365 gm of B₂H₆ ,

Molecular mass of B₂H₆ is 27.66 gm

so no. of moles = 365/27.66 = 13.19

To produce 2 moles of B₂H₆ , 4 moles of BCl₃ is needed , therefore to make 13.19 moles of  B₂H₆ we will need following moles of BCl₃

$\rm \dfrac{4 \times 13.19}{2} = 26.39 \; moles$

Therefore 26.39 moles of BCl₃ will be required to produce 365 grams of B₂H₆.

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