TiO2(s) + H2(g) Ti(s) + H2O(g)
(a) Consider the unbalanced equation above. What is the maximum mass of titanium that can be produced when 2.65 g of TiO2 and 3.40 g H2 react?
= ?g Ti
(b)Assume the reaction in part (a) goes to completion. Calculate the mass of excess reactant.
=?g
There is a missing arrow --> in your reaction.
Your first step should be to balance the reaction. You will twice as many H2's as O2's to remove the O atoms from TiO2
Someone be glad to critique your work.
67g
To answer these questions, we'll need to balance the chemical equation first.
(a) Balancing the equation:
TiO2(s) + H2(g) -> Ti(s) + H2O(g)
To balance the equation, we need the same number of atoms on both sides. To do this, we start by counting the atoms present on each side of the equation:
On the left side:
Ti: 1 Ti atom
O: 2 O atoms
On the right side:
Ti: 1 Ti atom
H: 2 H atoms
O: 1 O atom
Now, we can balance the equation by adjusting the coefficients (numbers in front of each compound):
TiO2(s) + 2 H2(g) -> Ti(s) + 2 H2O(g)
Now, the equation is balanced.
(b) Now that we have a balanced equation, we can determine the limiting reactant and the excess reactant.
1. Determine the moles of each reactant:
To find the moles of each reactant, we'll use the molar masses of TiO2 and H2.
Molar mass of TiO2 = 79.87 g/mol
Molar mass of H2 = 2.02 g/mol
Moles of TiO2 = mass of TiO2 / molar mass of TiO2
Moles of TiO2 = 2.65 g / 79.87 g/mol = 0.0332 mol
Moles of H2 = mass of H2 / molar mass of H2
Moles of H2 = 3.40 g / 2.02 g/mol = 1.6832 mol
2. Calculate the stoichiometric ratio:
From the balanced equation, we can see that the ratio of the coefficients between TiO2 and Ti is 1:1. This means that for every 1 mole of TiO2, we obtain 1 mole of Ti. Therefore, the number of moles of Ti produced will be the same as the number of moles of TiO2 used.
Moles of Ti = 0.0332 mol
3. Convert moles of Ti to grams:
Mass of Ti = Moles of Ti * molar mass of Ti
Mass of Ti = 0.0332 mol * 47.87 g/mol = 1.59 g
So, the maximum mass of titanium that can be produced when 2.65 g of TiO2 and 3.40 g H2 react is 1.59 g.
4. Calculate the mass of excess reactant:
To find the mass of excess reactant, we'll calculate the stoichiometric ratio between H2 and Ti.
From the balanced equation, we can see that the ratio of the coefficients between H2 and Ti is 2:1. This means that for every 2 moles of H2, we need 1 mole of Ti. Therefore, to find the moles of H2 required to react with the given moles of TiO2, we need to use the stoichiometric ratio.
Moles of H2 required = 0.5 * Moles of Ti = 0.5 * 0.0332 mol = 0.0166 mol
Moles of H2 present = 1.6832 mol
Excess moles of H2 = Moles of H2 present - Moles of H2 required
Excess moles of H2 = 1.6832 mol - 0.0166 mol = 1.6666 mol
Mass of excess H2 = Excess moles of H2 * molar mass of H2
Mass of excess H2 = 1.6666 mol * 2.02 g/mol = 3.365 g
So, the mass of excess reactant (H2) is 3.365 g.