Chapter 15.2, Problem 13R

To determine

The amount of bismuth fluoride formed.

Answer to Problem 13R

  $531.9$ gramsbismuth fluoride formed.

Explanation of Solution

Given information:

The amount of bismuth is $417.96$ gramsand the amount of fluoride is $113.99$ grams.

Formula used:

Consider the reaction between bismuth and fluorine.

  $2\text{Bi}+3{\text{F}}_2\stackrel{}{\to }2{\text{BiF}}_3$

Calculation:

The molar mass of fluorine is $18.99$ grams per mole and the molar mass of bismuth is $208.98$ grams per mole.

Since the molecular formula of bismuth fluoride is ${\text{BiF}}_3$ . Calculate the molar mass of bismuth fluoride.

  $\begin{array}{c}\text{Molar}\text{mass}=208.98+3\left(18.99\right)\\ =265.95\text{g/mol}\end{array}$

Since, the reactants are bismuth and fluorine. The amount of bismuth fluoride is based on the starting mass fluoride. Calculate the mole product of fluoride.

  $\text{Fluoride}=\left(113.99\text{g}{\text{F}}_2\times 1\text{mol}{\text{F}}_2/38\text{g}{\text{F}}_2\times 2\text{mol}{\text{BiF}}_3/3\text{mol}{\text{F}}_2\right)=1.999\text{mol}{\text{BiF}}_3$

Calculate the mass of bismuth fluoride.

  $1.999\text{mol}{\text{BiF}}_3\times 265.95\text{g/mol}{\text{BiF}}_3/1\text{mol}{\text{BiF}}_3=531.91\text{g}{\text{BiF}}_3$

Therefore, the amount of bismuth fluoride formed is $531.91$ grams.

Conclusion:

Hence, the amount of bismuth fluoride formed is $531.91$ grams.