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 to produce a given volume of dry N2 gas, we need to use the ideal gas law equation. The ideal gas law equation is given by:
PV = nRT
Where:
P = pressure in atmospheres
V = volume in liters
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, we need to convert the given volume from milliliters to liters. We know that 1 liter is equal to 1000 milliliters, so 448.2 mL is equivalent to 448.2 / 1000 = 0.4482 liters.
Next, we need to convert the temperature from Celsius to Kelvin. The Kelvin temperature scale is obtained by adding 273.15 to the Celsius temperature. So, 22.3 °C + 273.15 = 295.45 K.
Now, we can plug in the values into the ideal gas law equation to find the number of moles of N2 gas:
PV = nRT
(777.3 torr) * (0.4482 L) = n * (0.0821 L·atm/mol·K) * (295.45 K)
To cancel out the units and find n, we can rearrange the equation:
n = (PV) / (RT)
Now, let's calculate the value of n:
n = (777.3 torr * 0.4482 L) / (0.0821 L·atm/mol·K * 295.45 K)
n ≈ 15.15 moles
According to the balanced chemical equation:
3 KOBr + 2 NH3 → N2 + 3 KBr + 3 H2O
We see that 2 moles of NH3 will produce 1 mole of N2. Therefore, we need half the number of moles of NH3 to produce the same amount of N2, which is 15.15 / 2 = 7.575 moles.
Now, we need to calculate the molar mass of KOBr. The molar mass of K is 39.10 g/mol, O is 16.00 g/mol, and Br is 79.90 g/mol. Adding these values together gives a molar mass of KOBr as 39.10 + 16.00 + 79.90 = 134.00 g/mol.
To calculate the minimum mass of KOBr required, we can multiply the number of moles (7.575) by the molar mass (134.00 g/mol):
Mass of KOBr = 7.575 moles * 134.00 g/mol
Mass of KOBr ≈ 1016.19 grams
Therefore, the minimum number of grams of KOBr required to produce 448.2 mL of dry N2 gas, when measured at 22.3 °C and 777.3 torr, is approximately 1016.19 grams.