1. What amount of Potassium Chloride forms when 5.25 litres of Chloride gas at 0.95 atm and 293K reacts with 17.0g of Potassium.

Here is a worked example. Just follow the steps. http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine the amount of Potassium Chloride formed, we need to balance the chemical equation for the reaction between Chloride gas (Cl2) and Potassium (K). The balanced equation is:

2 K + Cl2 -> 2 KCl

From the balanced equation, we can see that 2 moles of Potassium (K) react with 1 mole of Chlorine gas (Cl2) to produce 2 moles of Potassium Chloride (KCl).

First, let's find the number of moles of Potassium (K) using its molar mass:

Molar mass of Potassium (K) = 39.10 g/mol

Number of moles of Potassium (K) = mass of Potassium (K) / molar mass of Potassium (K)

Number of moles of Potassium (K) = 17.0 g / 39.10 g/mol = 0.435 mol

Since the stoichiometric ratio between Potassium (K) and Potassium Chloride (KCl) is 2:2, 0.435 moles of Potassium (K) will produce the same amount of moles of Potassium Chloride (KCl).

Therefore, the amount (in moles) of Potassium Chloride (KCl) formed is 0.435 mol.

Next, we can calculate the volume of Chloride gas (Cl2) at the given conditions (0.95 atm and 293K) using the Ideal Gas Law equation:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/K·mol)
T = temperature (in Kelvin)

Rearranging the equation to solve for V:

V = (nRT) / P

Plugging in the values:

V = (0.435 mol * 0.0821 L·atm/K·mol * 293K) / 0.95 atm

V ≈ 11.36 L

Therefore, 5.25 liters of Chloride gas at 0.95 atm and 293K will result in the formation of approximately 11.36 L of Potassium Chloride (KCl).

To determine the amount of Potassium Chloride formed, we need to first write a balanced chemical equation for the reaction between Chloride gas (Cl₂) and Potassium (K).

The balanced equation is:

2 K + Cl₂ -> 2 KCl

This equation tells us that for every 2 moles of Potassium (K), we need 1 mole of Chloride gas (Cl₂) to form 2 moles of Potassium Chloride (KCl).

To find the amount of Potassium Chloride formed, we will use the concept of stoichiometry.

Step 1: Convert the given volume of Chloride gas (Cl₂) to moles. We will use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

First, convert the volume from liters to moles using the given pressure and temperature:
PV = nRT
(0.95 atm) * (5.25 L) = n * (0.0821 L·atm/(mol·K)) * (293 K)
Solving for n, we find n ≈ 1.274 moles.

Step 2: Use the stoichiometry of the balanced chemical equation to find the moles of Potassium Chloride formed.

From the balanced equation, we know that 2 moles of Potassium (K) react to form 2 moles of Potassium Chloride (KCl). Therefore, 1.274 moles of Chloride gas (Cl₂) will react to form the same number of moles of Potassium Chloride (KCl).

Step 3: Convert moles of Potassium Chloride (KCl) to grams using the molar mass of Potassium Chloride (KCl).

The molar mass of Potassium Chloride (KCl) is K: 39.10 g/mol + Cl: 35.45 g/mol = 74.55 g/mol.

To find the mass of Potassium Chloride formed, we multiply the moles of Potassium Chloride by its molar mass:

Mass of Potassium Chloride (KCl) = 1.274 moles * 74.55 g/mol ≈ 95.008 g.

Therefore, approximately 95.008 grams of Potassium Chloride (KCl) will be formed when 5.25 liters of Chloride gas at 0.95 atm and 293 K reacts with 17.0 grams of Potassium (K).