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.