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How many g of #B_2H_6# will react with 3.00 mol of #O_2#?

Answer 1

Jeremiah Adams

Approx. #28*g#

A stoichiometrically balanced equation is required:

#B_2H_6(g) + 3O_2(g) rarr 2B(OH)_3(s)#

Is everything in balance?

And thus we need 1 equiv diborane, i.e. #2xx13.83*g#.

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Answer 2

Noah Aldrich

The balanced chemical equation for the reaction between B₂H₆ and O₂ is:

2 B₂H₆ + 6 O₂ -> 4 B₂O₃ + 6 H₂O

From the equation, it is evident that 2 moles of B₂H₆ react with 6 moles of O₂.

To find out how many grams of B₂H₆ will react with 3.00 mol of O₂, we use stoichiometry.

Given that 3.00 mol of O₂ is provided, we can set up a proportion:

3.00 mol O₂ / 6 mol O₂ = x mol B₂H₆ / 2 mol B₂H₆

Solving for x:

x = (3.00 mol O₂ / 6 mol O₂) * 2 mol B₂H₆ x = 1.00 mol B₂H₆

Now, we need to convert moles of B₂H₆ to grams using its molar mass.

The molar mass of B₂H₆ is approximately 27.67 g/mol.

So, 1.00 mol of B₂H₆ is equivalent to:

1.00 mol * 27.67 g/mol = 27.67 grams of B₂H₆.

Therefore, 27.67 grams of B₂H₆ will react with 3.00 mol of O₂.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.