Chemistry: Limiting Reactants Help?

Question

What is the maximum number of grams of B2H6(g) that can be prepared from 2.650g of NaBH4(s) and 4.560g of BF3(g)?

Equation for Reaction:

3 NaBH4(s) 4 BF3(g) --> 3 NaBF4(s) 2 B2H6(g)

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Thank you so much.

Answer 1

from 2.65 g of NaBH4

     2.65g NaBH4[37.83253g/mol NaBH4 / 1g NaBH4] [2 mol B2H6 / 3 mol NaBH4] [ ? g B2H6 / 27.66964 g/mol B2H6] = 2. 415556 g of B2H6

from 4.56 g BF3

     4.56g BF3[67.80621 g/mol BF3 / 1g BF3] [ 2 mol B2H6 / 4 mol BF3] [?g B2H6 / 27.66964 g/mo B2H6] = 5.587285

Therefore the limitting reactant is NaBH4 since it yields the lower g of B2H6 which is 2.415556 g.

and therefore the maximum number of grams of B2H6 that can be prepared based on reactions is 2.415556{theoretical yield}

Answer 2

Moles NaBH4 = 2.650 g / 37.8328 g/mol =0.07004

moles BF3 = 4.560 g / 67.8062 g/mol = 0.06725

the ratio between reactants is 3 : 4

so BF3 is the limiting reactant ( 4 x 0.07004 / 3 = 0.09339 moles of BF3 are needed to react with 0.07004 moles of NaBH4)

the ratio between BF3 and B2H6 is 4 : 2

Moles B2H6 = 0.06725 x 2 / 4 = 0.03363

Mass B2H6 = 0.03363 mol x 27.67 g/mol =0.9305 g