# Calculate the mass of 1.50 L of CH4 at STP

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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