Question

For the following reaction, 12.8 grams of sulfur are allowed to react with 26.2 grams of carbon monoxide.

sulfur(s) + carbon monoxide(g) → sulfur dioxide(g) +carbon(s)

What is the maximum amount of sulfur dioxide that can be formed?  grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete?  grams

Answer 1

As, Number of moles = Given mass in g / Gram molar mass

So,

Number of moles of S = Mass of S / Gram molar mass of S

= 12.8 g / 32 g/mol

= 0.40 mol

Similarly,

Number of moles of CO = Mass of CO / Gram molar mass of CO

= 26.2 g / 28 g/mol

= 0.9357 mol

Balanced chemical equation is :

S (s) + 2CO (g) ------------------> SO₂ (g) + 2C (s)

From the Balanced chemical equation :

2 mol of CO reacts with = 1 mol of S

So,

0.9357 mol of S reacts with = 0.9357 / 2 mol of S

= 0.4678 mol of S

Because, Given number of moles of S are less i.e. only 0.40 mol as compare to required i.e. 0.4678 mol , therefore Sulfur will be the Limiting reagent (i.e. the reactant which is 100 % consumed in the reaction).

Limiting reagent = Sulfur

and

CO gas will be the excessive reagent (i.e. the reactant whose amount left in the solution)

Now,

From ICF table :

........................S (s) ................+................ 2CO (g) ------------------> SO₂ (g).................. +.................. 2C (s)

Initial ...............0.40 mol...............................0.9357 mol ...................0.0 mol.......................................0.0 mol

Change ...........-0.40 mol.............................-2x0.40 mol....................+0.40 mol...........................+2x0.40 mol

Final..................0.0 mol................................0.1357 mol......................0.40 mol.......................................0.40 mol

Moles of SO₂ gas formed = 0.40 mol

Mass of SO₂ gas formed = 0.40 mol x 64 g/mol

= 25.6 g

Maximum Mass of SO2 gas formed = 25.6 g

Now, The moles of excessive reagent i.e. CO remains = 0.1357 mol

So,

Mass of excessive reagent remains in the solution = Moles x Gram molar mass

= 0.1357 mol x 28 g/mol

= 3.80 g  

So,

Mass of excessive reagent i.e. CO left = 3.80 g