Question

Consider the reaction between TiO2 and C:

Answer 1

Answer:

TiO2 = Nothing ledt

C = 7g

Ti = 5.98g

CO =3.5g

Explanation:

TiO2(s) +2C(s) → Ti(s) +CO(g)

Initial reactant masses of 10g

This is a limiting reactant problem. We have to find the limiting reactant; this is the reactant that determines the amount of product formed.

From the stoichiometry of the reaction;

1 mol of TiO2 reacts completely with 2 moles of C

Converting the starting masses to molres, we have;

Moles = mass /  molar mass

For TiO2

Moles = 10g / 79.866 g/mol = 0.125 moles

For C

Moles = 10g / 12g/mol = 0.8333 moles

Time to find the limiting reactant;

If all the 0.125 moles of TiO2 were used, it would require 2* 0.125 = 0.250 moles of C. But we have 0.833 moles of C. This means C is in excess, hence TiO2 is our limiting reactant.

All of TiO2 would be used up.

Moles left of C = 0.8333 - 0.250 moles = 0.58333

Mass left = Moles * Molar mass = 0.5833 * 12 = 6.99 = 7g

Mass of Ti and Co formed;

Moles formed is 0.125 moles .

Mass of Ti = Moles * Molar mass = 0.125 * 47.867 = 5.98 g

Mass of CO = Moles * Molar mass = 0.125 * 28 = 3.5g