Chapter 11, Problem 82A

(a)

Interpretation Introduction

Interpretation:

For the given reaction, limiting reactant is to be determined.

Concept introduction:

Lithium react with bromine spontaneously to produce lithium bromide.

$2Li+Br2\to 2LiBr$

Lithium is a metal. Bromine is a gas and non-metal.

Answer to Problem 82A

Br₂ is limiting reactant.

Explanation of Solution

First convert the reactant in moles.

Using relation

$\text{Moles = }\frac{MassofCompound}{MolarMassofCompound}$

Molar mass of lithium is 6.93 g/mol

Molar mass of Bromine gas is 159.82 g/mol

$\begin{array}{l}\text{Moles of lithium = }\frac{25}{6.93}=3.60\\ \text{Moles of Bromine =}\frac{25}{159.82}=0.156\end{array}$

According to stoichiometry of reaction, 2 moles of lithium require 1 mole of bromine gas for complete reaction. And here Li is present in excess and left behind after completion of reaction. So, Br₂ is limiting reactant.

(b)

Interpretation Introduction

Interpretation:

Mass of lithium bromide produced to be determined

Concept introduction:

Lithium react with bromine spontaneously to produce lithium bromide.

$2Li+Br2\to 2LiBr$

Answer to Problem 82A

27.09 g of LiBr

Explanation of Solution

From reaction 1 mole of Br₂ form 2 moles of LiBr.

0.156 mole of Br₂ form 2 times the mole of LiBr.

= 0.312 mole of LiBr

Molar mass of LiBr is 86.85g/mol

Mass of LiBr formed is:

$\begin{array}{l} m=0.312\times 86.85\hfill \\ =27.09gofLiBr\hfill \end{array}$

(c)

Interpretation Introduction

Interpretation:

Excess reactant and excess mass to be determined.

Concept introduction:

Lithium react with bromine spontaneously to produce lithium bromide.

$2Li+Br2\to 2LiBr$

Answer to Problem 82A

3.288 mole of Li is in excess

Mass of excess Li = $22.78g$

Explanation of Solution

2 moles of Li will form 2 moles of LiBr

Thus, 0.312 mole of LiBr was formed from 0.312 mole of Li.

Also, 3.60 mole of lithium is present.

From that amount 0.312 mole reacts, hence remaining Li = 3.60 − 0.312 = 3.288 mole

3.288 mole of Li is in excess

Mass of excess Li =

$3.288\times 6.93=22.78g$