Potassium hypobromite, KOBr, converts ammonia to nitrogen by the following reaction.
3KOBr + 2NH3 N2 + 3KBr + 3H2O
To prepare 448.2 mL of dry N2, when measured at 22.3 °C and 777.3 torr, what is the minimum number of grams of KOBr required?
To determine the minimum number of grams of KOBr required, we need to use the ideal gas law equation to find the number of moles of nitrogen (N2) gas produced, and then use the balanced equation to establish the stoichiometric relationship between KOBr and N2.
1. Start by converting the given volume of N2 (448.2 mL) to moles using the ideal gas law equation:
PV = nRT
Where:
P = pressure (777.3 torr)
V = volume (448.2 mL)
n = number of moles
R = gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin (22.3 °C + 273.15 = 295.45 K)
Rearranging the equation to solve for moles (n):
n = PV / RT
Substituting the given values:
n = (777.3 torr * 0.4482 L) / (0.0821 L·atm/mol·K * 295.45 K)
n = 0.0131 moles of N2
2. Now, using the balanced equation provided, we can determine the stoichiometric ratio between KOBr and N2:
3 KOBr + 2 NH3 -> N2 + 3 KBr + 3 H2O
According to the balanced equation, 3 moles of KOBr produce 1 mole of N2. Therefore, we can establish the following ratio:
3 moles KOBr : 1 mole N2
3. Calculate the moles of KOBr required by multiplying the moles of N2 by the stoichiometric ratio:
moles of KOBr = (0.0131 moles N2) * (3 moles KOBr / 1 mole N2)
moles of KOBr = 0.0393 moles KOBr
4. Finally, we can determine the mass of KOBr required using the molar mass of KOBr (from the periodic table):
Molar mass of KOBr = atomic mass of K + atomic mass of O + atomic mass of Br
Molar mass of KOBr = (39.10 g/mol K) + (15.999 g/mol O) + (79.90 g/mol Br)
Molar mass of KOBr = 134.01 g/mol
Mass of KOBr = moles of KOBr * molar mass of KOBr
Mass of KOBr = (0.0393 moles KOBr) * (134.01 g/mol KOBr)
Mass of KOBr ≈ 5.26 grams
Therefore, the minimum number of grams of KOBr required to produce 448.2 mL of dry N2 at 22.3 °C and 777.3 torr is approximately 5.26 grams.