Question

In a combination reaction, 1.62 g of lithium is mixed with 6.50 g of oxygen.

(a) Which reactant is present in excess?

(b) How many moles of product are formed?

(c) After reaction, how many grams of each reactant and product are present?

Short answer

(a) O₂ is present in excess.

(b) Moles of LiO₂ (product) are 0.1165mol

(c) After reaction, there are 0g Li, 4.636G O₂ and 3.481G LiO₂ .

Step-by-step solution

Write the balanced equation

A combination reaction of lithium with oxygen forms lithium oxide. The balanced chemical equation is:

$4Li\left(s\right)+O_2\left(g\right)\to 2L{i}_{2}O\left(s\right)$

Calculation of moles of Li

According to the question,

Mass of Li = 1.62 g

Molar mass of Li = 6.941 g/mol

Known,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Thus,

$\begin{array}{rcl}numberofmolesofLi& =& \frac{1.62}{6.941}\left(mol\right)\\ & =& 0.233\left(mol\right)\end{array}$

Hence, number of moles of Li is 0.233 mol.

Calculation of moles of   when Li is limiting reagent

According to the balanced equation, 4 moles of Li produces 2 moles LiO₂.

$4Li\left(s\right)+O_2\left(g\right)\to 2L{i}_{2}O\left(s\right)$

Now,

$\begin{array}{rcl}numberofmolesofL{i}_{2}O& =& 0.233\times \frac{2}{4}\left(mol\right)\\ & =& 0.1165\left(mol\right)\end{array}$

Hence, moles of LiO₂ when Li is limiting reagent are 0.1165 mol .

Calculation of moles of O2

According to the question,

Mass of O₂ = 6.50 g

Molecular mass of O₂ = 32 g/mol

Since,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Thus,

$\begin{array}{rcl}numberofmolesof{O}_2& =& \frac{6.50}{32}\left(mol\right)\\ & =& 0.203\left(mol\right)\end{array}$

Hence, moles of O₂ are 0.203mol.

Calculation of moles of Li2O when O2 is limiting reagent

According to the balanced equation,1 mole of O₂ produces 2 moles LiO₂.

$4Li\left(s\right)+O_2\left(g\right)\to 2L{i}_{2}O\left(s\right)$

Now,

$\begin{array}{rcl}numberofmolesofL{i}_{2}O& =& 0.203\times \frac{2}{1}\left(mol\right)\\ & =& 0.406\left(mol\right)\end{array}$

Hence, moles of LiO₂ when O₂ is limiting reagent are 0.406 mol.

Determine the Limiting reagent

In the given reaction,

$4Li\left(s\right)+O_2\left(g\right)\to 2L{i}_{2}O\left(s\right)$

Li is the limiting reagent in this reaction as moles of LiO₂ is less in terms of Li.

Hence, O₂ is present in excess.

Calculation of moles of Li2O (product)

According to the balanced equation, 4 moles of Li produces 2 moles of LiO₂.

$4Li\left(s\right)+O_2\left(g\right)\to 2L{i}_{2}O\left(s\right)$

Now, moles of LiO₂ is:

$\begin{array}{rcl}numberofmolesofLi{O}_2& =& 0.233\times \frac{2}{4}\left(mol\right)\\ & =& 0.1165\left(mol\right)\end{array}$

Hence, moles of LiO₂ (product) are 0.1165 mol.

Calculation of mass of Li (reactant)

As Li is the limiting reagent; so after reaction there will be no Li.

Calculation of moles of O2 reacted

From the balanced equation, 4 moles of Li reacts with 1 mole O₂

$4Li\left(s\right)+O_2\left(g\right)\to 2L{i}_{2}O\left(s\right)$

Hence,

$\begin{array}{rcl}numberofmolesof{O}_{2}reacted& =& 0.233\times \frac{1}{4}\left(mol\right)\\ & =& 0.05825\left(mol\right)\end{array}$

Hence, moles of O₂ reacted is 0.05825 mol.

Calculation of mass of O2 reacted

Since,

mass = moles x mass (molar)

Molar mass of O₂ is 32g/mol.

Now,

$\begin{array}{rcl}Massof{O}_2& =& 0.05825\times 32\left(g\right)\\ & =& 1.864\left(g\right)\end{array}$

Hence, mass of O₂ reacted is 1.864 g.

Calculation of mass of O2 left (reactant)

Mass of O₂ = mass(initial) - mass(reacted)

= (6.50 - 1.864)(g)

= 4.636(g)

Hence, mass of O₂ left is 4.636g.

Calculation of mass of Li2O (product)

Since,

mass = moles x mass (molar)

Molar mass of LiO₂ is 29.882g/mol.

Now, mass of LiO₂ is:

$\begin{array}{rcl}MassofLi{O}_2& =& 0.1165\times 29.882\left(g\right)\\ & =& 3.481\left(g\right)\end{array}$

Hence, mass of LiO₂ is 3.481g.