the equation is TiCl4 + O2 (yields) TiO2 + 2Cl2.

How much titanium(IV) chloride must react with excess oxygen to prepare 175 grams of the pigment? The reaction is found to be 85% efficient? (%yield)??

A worked example, in detail.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine how much titanium(IV) chloride (TiCl4) must react with excess oxygen (O2) to prepare 175 grams of the pigment, we need to calculate the stoichiometry of the reaction.

From the balanced equation:
1 mol TiCl4 reacts with 1 mol O2 to produce 1 mol TiO2 and 2 mol Cl2.

Now, let's calculate the molar mass of the compound:
- TiCl4: 1 atom of titanium (Ti) with a molar mass of 47.867 g/mol and 4 atoms of chlorine (Cl) with a molar mass of 35.453 g/mol, giving a total molar mass of 189.679 g/mol.

To find the number of moles of TiCl4 required to produce 175 grams of TiO2, we can use the molar mass of TiCl4:
175 g ÷ 189.679 g/mol = 0.9231 mol TiO2

Since the reaction is found to be 85% efficient, the actual yield of TiO2 will be 85% of the theoretical yield.

Actual yield = efficiency × theoretical yield
Actual yield = 0.85 × 0.9231 mol = 0.7847 mol TiO2

According to the stoichiometry of the balanced equation, 1 mol of TiO2 is produced from 1 mol of TiCl4. Therefore, the same amount of TiCl4 is required.

Therefore, you will need 0.7847 mol of TiCl4 to prepare 175 grams of the pigment.

To determine the mass of TiCl4 required, we can use the molar mass of TiCl4:
0.7847 mol × 189.679 g/mol = 149.18 grams of TiCl4.

Hence, approximately 149.18 grams of titanium(IV) chloride must react with excess oxygen to prepare 175 grams of the pigment.

To determine how much titanium(IV) chloride (TiCl4) must react with excess oxygen (O2) to prepare 175 grams of the pigment, we need to calculate the stoichiometry of the reaction.

First, let's write down the balanced chemical equation:

TiCl4 + O2 -> TiO2 + 2Cl2

From the equation, we can see that 1 mole of TiCl4 reacts with 1 mole of O2 to produce 1 mole of TiO2 and 2 moles of Cl2.

Next, we need to calculate the molar mass of TiCl4:
Ti (titanium) = 47.9 g/mol
Cl (chloride) = 35.45 g/mol (there are four chlorides, so 4 x 35.45 = 141.8 g/mol)

Molar mass of TiCl4 = Ti + 4Cl = 47.9 + 141.8 = 189.7 g/mol

Now, we can calculate the number of moles of TiCl4 needed to produce 175 grams of TiO2, considering the reaction is 85% efficient:

Number of moles = Mass / Molar mass
Number of moles of TiCl4 = 175 g / 189.7 g/mol = 0.922 moles

However, since the reaction is only 85% efficient (85% yield), we need to account for the percent yield. To do this, we divide the number of moles by the percent yield:

Actual moles of TiCl4 = (Number of moles) / (Percent yield / 100)
Actual moles of TiCl4 = 0.922 moles / (0.85) = 1.08 moles (rounded to two decimal places)

Therefore, approximately 1.08 moles of TiCl4 would need to react with excess oxygen to prepare 175 grams of the pigment.