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Chemistry
Quentin
9 May, 19:51
Calculate the mass of 1.50 L of CH4 at STP
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Reginald Potter
9 May, 20:20
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Let's assume that CH₄ has ideal gas behavior.
Then we can use ideal gas formula,
PV = nRT
Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant (8.314 J mol⁻¹ K⁻¹) and T is temperature in Kelvin.
P = 1 atm = 101325 Pa
V = 1.50 L = 1.50 x 10⁻³ m³
n = ?
R = 8.314 J mol⁻¹ K⁻¹
T = 0 °C = 273 K
By substitution,
101325 Pa x 1.50 x 10⁻³ m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K
n = 0.0669 mol
Hence, moles of CH₄ = 0.0669 mol
Moles = mass / molar mass
Molar mass of CH₄ = 16 g mol⁻¹
Mass of CH₄ = moles x molar mass
= 0.0669 mol x 16 g mol⁻¹
= 1.0704 g
Hence, mass of CH₄ in 1.50 L at STP is 1.0704 g
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