What is the volume of 0.674 kg of acetylene, C2H2, at STP?
0.674g/molar mass = mols.
1 mol occupies 22.4L at STP.
To calculate the volume of a gas at STP (Standard Temperature and Pressure), we need to 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 in Kelvin.
First, we need to determine the number of moles of acetylene, C2H2. To do this, we can use the molar mass of acetylene to convert the mass to moles.
The molar mass of acetylene (C2H2) is calculated as follows:
(2 x atomic mass of carbon) + (2 x atomic mass of hydrogen)
= (2 x 12.01 g/mol) + (2 x 1.01 g/mol)
= 24.02 g/mol + 2.02 g/mol
= 26.04 g/mol
Now, we can calculate the number of moles using the given mass of acetylene (0.674 kg):
Number of moles = mass / molar mass
= 0.674 kg / 26.04 g/mol
= 674 g / 26.04 g/mol
≈ 25.87 mol
Since STP is defined as 1 atmosphere of pressure and 273.15 K (0 degrees Celsius), we can substitute these values into the ideal gas law equation.
P = 1 atm
V = ?
n = 25.87 mol
R = 0.0821 L•atm/(mol•K)
T = 273.15 K
Rearranging the equation, we get:
V = (n * R * T) / P
Substituting the given values, we can calculate the volume (V):
V = (25.87 mol * 0.0821 L•atm/(mol•K) * 273.15 K) / 1 atm
≈ 592.7 L
Therefore, the volume of 0.674 kg of acetylene, C2H2, at STP is approximately 592.7 liters.