Question

What mass of Ag₂O in grams is required to form 388 mL of oxygen gas at P = 734 mmHg?"

A) 0.134 g
B) 0.268 g
C) 0.402 g
D) 0.536 g

Answer 1

To find the mass of Ag2O required to form a certain volume of oxygen gas at a given pressure, we can use the ideal gas law equation. We first calculate the number of moles of Ag2O using the ideal gas law equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. Then, we multiply the number of moles by the molar mass of Ag2O to find the mass in grams.

Step-by-step explanation:

To find the mass of Ag2O required, we need to use the ideal gas law equation PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. The equation can be rearranged to solve for n, the number of moles. Once we have the number of moles, we can use the molar mass of Ag2O to find the mass in grams. Let's go through the steps:

Step 1: Given: P = 734 mmHg, V = 388 mL, T = ?, n = ?

Step 2: Convert pressure to atm - P = 734 mmHg = 0.966 atm

Step 3: Convert volume to liters - V = 388 mL = 0.388 L

Step 4: Use the ideal gas law equation to solve for n: PV = nRT

0.966 atm * 0.388 L = n * (0.0821 L·atm/(mol·K)) * T

n = (0.966 atm * 0.388 L) / (0.0821 L·atm/(mol·K) * T)

Step 5: Use the molar mass of Ag2O to find the mass in grams - molar mass of Ag2O = 231.74 g/mol

mass = n * molar mass of Ag2O

Now, you can substitute the values of n and the molar mass of Ag2O into the equation to find the mass in grams.