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Question

The question says Diborane(b2H6) is widley used in the synthesis of organic compounds. Diborane itself is made by the reaction 2NaBH4+I2=B2H2+2NaI+h2. If 6.3g of NaBH4 are reacted with excess I2, how many grams of diborane could theoretically be isolated? If 1.9 g of diborane is actually produced, what is the percent yield of the reaction?
Does anyone know hw to figure this out?

Answer 1

First start with the given mass of NaBH4. Since this is the limiting reactant, all of it will react, so the moles of NaBH4 = 2 times the moles of product formed (as can be seen by the mole ratio).

Figure out the moles of NaBH4 by dividing this mass by the molar mass (molar mass of Na + molar mass of B + 4 times the molar mass of H). This will give you the moles of NaBH4, and for every mole of this that reacts, 1/2 mole of B2H2 and 1/2 mole of H2 forms (so 1/2 mole of B2H6 can be isolated).
Find the mass of diborane by multiplying the molar mass of diborane (twice the molar mass of B + six times the molar mass of H) by 1/2 the moles of NaBH4 that you found reacted. This will give you the first answer.

% yield = mass produced divided by theoretical mass, so divide 1.9 g by the mass you found and multiply by 100 to get a percent.

Good Luck!

Answer 2

first we need to figure out how many moles of NaBH4 is in 6.3 grams. to get that we need to divide 6.3 by the molecular weights of Na+B+4H. now that we know how much we started with(in moles) then we can figure out how much we are going to get. we learned from the formula that it takes 2 moles of NaBH4 in order to get 1 mole of B2H2. so we take our initial moles and divide it by two. now we have the moles of B2H2 produced. if we multiply it by the molar mass of the compound then we will know how many grams we got.
to find the percent yield just divide the Actual Yield by the Theoretical Yield

Answer 3

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Answer 4

2NaBH4 + I2 --> B2H6 + 2NaI + H2

This reaction says that 2 moles NaBH4 yield 1 mole B2H6.

Molecular weight of NaBH4 = 37.83 g/mol, and
molecular weight of B2H6 = 27.67 g/mol.

Therefore, 2 * 37.83 g NaBH4 yields 1 * 27.67 g B2H6

So, 6.3 g NaBH4 yields 6.3 * (1 * 27.67) / (2 * 37.83) g B2H6,
which is 2.3 g.

If only 1.9 g is produced, then yield = (1.9 / 2.3) * 100 = 82.6%.