The volume of 17.5 grams of SO2(g) at STP would be
find volume in liters
PV=nRT
First find moles(n) from grams/ divide number of grams given by molar mass of SO2- 17.5/64.062
Then plug it into formula with STP
T=273K P= 1 atm R is always 0.0821 with atm’s
(1)(x)=(0.27317)(0.0821)(273)
=6.12 L
V=nRT/P n= 17.5/molmassSO2 watch units
To find the volume of 17.5 grams of SO2(g) at STP (Standard Temperature and Pressure), we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (STP is 1 atm)
V = volume (what we want to find)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (STP is 273.15 K)
First, we need to calculate the number of moles of SO2. To do that, we can use the molar mass of SO2, which is the sum of the atomic masses of sulfur (32.06 g/mol) and oxygen (2 * 16.00 g/mol).
Molar mass of SO2 = (32.06 g/mol) + (2 * 16.00 g/mol) = 64.06 g/mol
Next, we can use the equation:
n = mass / molar mass
n = 17.5 g / 64.06 g/mol ≈ 0.273 mol
Now, we can substitute the known values into the ideal gas law equation:
(1 atm) * V = (0.273 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K)
Simplifying the equation:
V = (0.273 mol) * (0.0821 L·atm/(mol·K)) * (273.15 K) / (1 atm)
V ≈ 6.66 L
Therefore, the volume of 17.5 grams of SO2(g) at STP would be approximately 6.66 liters.