Consider the reaction between TiO2 and C: TiO2(s) +2C(s) → Ti(s) +CO(g)

A reaction vessel initially contains 10.0g each of the reactants. Calculate the masses of TiO2, C, Ti, and CO that will be in the reaction vessel after the reactants have reacted as much as possible

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

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.

the right answer is c i just did the question.