Assume that 7.2 L of I2 are produced at STP according to the following balanced equation:
2KI+Cl - 2KCl+ I2
How many grams of KI (MOLAR MASS=166.002G/MOL) reacted
See your post above.
To find the number of grams of KI that reacted, we need to use stoichiometry – the relationship between the number of moles of reactants and products in a chemical equation.
First, let's determine the molar ratio between KI and I2 using the balanced equation:
2 KI + Cl2 → 2 KCl + I2
For every 2 moles of KI, we produce 1 mole of I2. Therefore, the molar ratio is 2:1.
Next, we'll find the number of moles of I2 produced. Since the volume is given at STP, we can use the ideal gas law:
PV = nRT
At STP (Standard Temperature and Pressure):
P = 1 atm
V = 7.2 L
T = 273 K
R = 0.0821 L.atm/mol.K
n = PV / RT
n = (1 atm) * (7.2 L) / (0.0821 L.atm/mol.K * 273 K)
Simplifying the units, we get:
n ≈ 0.333 mol
Now, using the molar ratio, we can calculate the number of moles of KI reacted:
n(KI) = n(I2) * (2 moles KI / 1 mole I2)
n(KI) = 0.333 mol * 2
n(KI) ≈ 0.666 mol
Finally, we can calculate the mass of KI that reacted using the molar mass (166.002 g/mol):
mass(KI) = n(KI) * molar mass(KI)
mass(KI) = 0.666 mol * 166.002 g/mol
mass(KI) ≈ 110 g
Therefore, approximately 110 grams of KI reacted.