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

The answer to this problem is. 67