How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal?
Here is a sample stoichiometry problem I've posted. Just follow the steps. The equation you want is
F2 + 2K = 2KF
http://www.jiskha.com/science/chemistry/stoichiometry.html
To determine the number of liters of fluorine gas that will react with 23.5 grams of potassium metal, we need to use balanced chemical equations and the concept of stoichiometry.
The balanced chemical equation for the reaction between fluorine gas (F2) and potassium metal (K) is:
2K + F2 -> 2KF
From the equation, we can see that 2 moles of potassium react with 1 mole of fluorine gas to produce 2 moles of potassium fluoride.
First, we need to calculate the number of moles of potassium (K) using its molar mass. The molar mass of potassium is 39.10 g/mol.
Number of moles of potassium (K) = mass of potassium (g) / molar mass of potassium (g/mol)
Number of moles of potassium (K) = 23.5 g / 39.10 g/mol
Number of moles of potassium (K) ≈ 0.599 mol
According to the balanced equation, the molar ratio between potassium (K) and fluorine gas (F2) is 2:1. Therefore, the number of moles of fluorine gas required is half the number of moles of potassium.
Number of moles of fluorine gas (F2) = 0.599 mol / 2
Number of moles of fluorine gas (F2) ≈ 0.300 mol
To convert the number of moles of fluorine gas to liters, we can use the Ideal Gas Law, which states that at standard temperature and pressure (STP: 0°C or 273.15 K and 1 atm), one mole of ideal gas occupies 22.4 liters of volume.
Number of liters of fluorine gas (F2) = number of moles of fluorine gas (F2) × 22.4 L/mol
Number of liters of fluorine gas (F2) ≈ 0.300 mol × 22.4 L/mol
Number of liters of fluorine gas (F2) ≈ 6.72 L
Therefore, approximately 6.72 liters of fluorine gas will react with 23.5 grams of potassium metal at standard temperature and pressure.