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
