Answer:
The number of liters of oxygen gas (O2) at STP is approximately 126.14 liters.
The number of molecules of bromine (Br2) produced is approximately 3.396 x 10^24 molecules.
The number of moles of titanium(IV) fluoride (TiF4) produced is approximately 1.878 moles.
The grams of BrF3 required for the complete reaction is approximately 889.8 grams.
Step-by-step explanation:
To find the quantities of different substances involved in the reaction between 150.0 grams of TiO2 and BrF3, we'll first need to write and balance the chemical equation for the reaction.
The balanced chemical equation for the reaction between TiO2 and BrF3 is as follows:
TiO2 + 3 BrF3 → TiF4 + 3 O2 + Br2
From the balanced equation, we can see that:
1 mole of TiO2 reacts with 3 moles of BrF3 to produce 1 mole of TiF4, 3 moles of O2, and 1 mole of Br2.
Step 1:
Convert grams of TiO2 to moles of TiO2.
Molar mass of TiO2 (Titanium Dioxide) = 47.88 g/mol + 15.999 g/mol * 2 = 79.87 g/mol
Number of moles of TiO2 = 150.0 g / 79.87 g/mol ≈ 1.878 moles
Step 2:
Determine the limiting reactant.
To find the limiting reactant, we need to compare the number of moles of TiO2 and BrF3 and see which one is present in a lower amount. According to the balanced equation, 1 mole of TiO2 requires 3 moles of BrF3.
Number of moles of BrF3 required = 1.878 moles of TiO2 * (3 moles of BrF3 / 1 mole of TiO2) ≈ 5.634 moles of BrF3
Since we have more moles of BrF3 (5.634 moles) than what is required (1.878 moles), TiO2 is the limiting reactant.
Step 3:
Calculate the number of moles of TiF4 and O2 produced.
From the balanced equation, we can see that 1 mole of TiO2 produces 1 mole of TiF4 and 3 moles of O2.
Number of moles of TiF4 produced = 1.878 moles of TiO2 * (1 mole of TiF4 / 1 mole of TiO2) ≈ 1.878 moles of TiF4
Number of moles of O2 produced = 1.878 moles of TiO2 * (3 moles of O2 / 1 mole of TiO2) ≈ 5.634 moles of O2
Step 4:
Calculate the number of molecules of bromine (Br2) produced.
Since 1 mole of Br2 contains Avogadro's number (6.022 x 10^23) of molecules, we can calculate the number of molecules of Br2 produced.
Number of molecules of Br2 = 5.634 moles of Br2 * (6.022 x 10^23 molecules per mole) ≈ 3.396 x 10^24 molecules of Br2
Step 5:
Calculate the volume of oxygen gas (O2) at STP (Standard Temperature and Pressure).
STP conditions: T = 273.15 K and P = 1 atm
1 mole of any gas occupies 22.4 liters at STP. So, the number of liters of O2 gas produced at STP is:
Volume of O2 gas at STP = 5.634 moles of O2 * 22.4 L/mol ≈ 126.14 liters of O2
Step 6:
Calculate the grams of BrF3 required for a complete reaction.
From the balanced equation, we can see that 1 mole of TiO2 reacts with 3 moles of BrF3.
Number of moles of BrF3 required = 1.878 moles of TiO2 * (3 moles of BrF3 / 1 mole of TiO2) ≈ 5.634 moles of BrF3
Molar mass of BrF3 (Bromine Trifluoride) = 79.904 g/mol + 18.998 g/mol * 3 = 157.89 g/mol
Mass of BrF3 required = 5.634 moles of BrF3 * 157.89 g/mol ≈ 889.8 grams of BrF3
So, for the complete reaction:
The number of liters of oxygen gas (O2) at STP is approximately 126.14 liters.
The number of molecules of bromine (Br2) produced is approximately 3.396 x 10^24 molecules.
The number of moles of titanium(IV) fluoride (TiF4) produced is approximately 1.878 moles.
The grams of BrF3 required for the complete reaction is approximately 889.8 grams.