what is the mass of sodium nitrate (85.00g/mol) that releases 2.55L of gas at STP?
To calculate the mass of sodium nitrate that releases 2.55L of gas at STP (Standard Temperature and Pressure), we need to use the ideal gas law equation and molar mass.
The ideal gas law equation is:
PV = nRT
where:
P = pressure (in atmospheres)
V = volume (in liters)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
At STP, the pressure (P) is 1 atmosphere and the temperature (T) is 273 K.
First, let's calculate the number of moles of the gas using the ideal gas law equation:
n = (PV) / (RT)
= (1 atm * 2.55 L) / (0.0821 L·atm/mol·K * 273 K)
≈ 0.111 moles
Now, we can find the mass using the molar mass of sodium nitrate:
Molar mass of sodium nitrate = 85.00 g/mol
Mass = moles × molar mass
= 0.111 moles × 85.00 g/mol
≈ 9.435 g
Therefore, the mass of sodium nitrate that releases 2.55L of gas at STP is approximately 9.435 grams.