Question

What madd of boron trifluoride can be made from 30.0g of fluorine

Answer 1

The question is incomplete , complete question is:

According to the balanced chemical equation, what mass of boron trifluoride can be made from 30.0 g of fluorine?

$2B(s) + 3F_2(g)\rightarrow 2BF_3(g)$

Answer:

35.78 grams of boron trifluoride can be made from 30.0g of fluorine.

Explanation:

$2B(s) + 3F_2(g)\rightarrow 2BF_3(g)$

Mass of fluorine gas = 30 .0 g

Moles of fluorine gas = $\frac{30.0 g}{38 g/mol}=0.7895 mol$

According to recation, 3 moles of fluorine gas gives 2 moles of boron trifluoride.

Then 0.7895 moles of fluorine gas will give:

$\frac{2}{3}\times 0.7895 mol=0.5263 mol$ of boron trifluoride.

Mass of 0.5263 moles of boron fluoride:

0.5263 mol × 68 g/mol = 35.78 g

35.78 grams of boron trifluoride can be made from 30.0g of fluorine.

Answer 2

Mass of boron trifluoride is 35.768 g.

Firstly we need to write the balanced chemical equation involved in this question:

$2B(s)+ 3F_2(g)$ → $2BF_3(g)$

It is given that:

Mass of fluorine= 30.0 g

So, we need to calculate moles of Fluorine

∵ Molar mass of fluorine= 38 g/mol

$n= \frac{m}{M} = \frac{30.0g}{38 gmol^{-1}} = 0.789 mol$

Now, for 3 moles of F₂ ; 2 moles of BF₃ are produced.

So, for 0.789 moles of F₂, number of moles of  BF₃ will be:

$\frac{2}{3} *0.789=0.526 \text{mol}$

Moles of  BF₃ =0.526 mol

∵ Molar mass of boron trifluoride =67.8 g/mol

Thus, mass of boron trifluoride can be calculated as:

$m=n*M= 0.526*67.8=35.662 g$

⇒Mass of boron trifluoride is 35.768 g.

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