How many liters of chlorine gas at 1.00 atm and 20ºC will be required to oxidize 167 grams of bromide ions to molecular bromine?
A. 13.5 L
B. 25.1 L
C. 101 L
D. 1.72 L
E. 50.3 L
http://www.jiskha.com/science/chemistry/stoichiometry.html
Pv=nRT
P=273
n=167g change to moles
R=22.414L/mol
T=293 K
solve for V
= 25.1 L
To determine the answer, we need to use the ideal gas law and stoichiometry.
Step 1: Convert the mass of bromide ions to moles.
To do this, we need to know the molar mass of bromide (Br-) ions. The molar mass of Br is approximately 79.90 g/mol. Therefore, 167 g of bromide ions is equal to:
167 g / (79.90 g/mol) = 2.087 mol
Step 2: Write the balanced chemical equation.
The balanced equation for the oxidation of bromide ions with chlorine gas is:
2 Br- (aq) + Cl2 (g) → 2 Cl- (aq) + Br2 (l)
This equation tells us that 2 moles of bromide ions react with 1 mole of chlorine gas to produce 1 mole of molecular bromine.
Step 3: Use stoichiometry to find the number of moles of chlorine gas required.
Since the ratio in the balanced equation is 2 moles of bromide ions to 1 mole of chlorine gas, we have:
2.087 mol Br- × (1 mol Cl2 / 2 mol Br-) = 1.044 mol Cl2
Step 4: Convert moles of chlorine gas to volume.
To do this, we need to know the molar volume of a gas at the given conditions. According to the ideal gas law, at 1.00 atm and 20ºC, the molar volume is approximately 24.5 L/mol.
Therefore, the volume of chlorine gas required is:
1.044 mol Cl2 × (24.5 L/mol) = 25.548 L
Rounded to the nearest tenth, the answer is 25.5 L.
The correct choice from the given options is B. 25.1 L.