Agas sample contains 4.00g of ch4 and 2.00g of he. what is the volume of the sample at stp?

As we know that 1 mole of any gas when acting perfectly / ideally occupies 22.4 L of volume.

It means that in order to calculate volume occupied both by CH₄ and He, we must calculate their moles.

Hence,
Moles are given as,
                                     Moles  =  Mass / Mass

Moles of CH₄;
                                     Mole-CH₄  =  4 g / 16 g.mol⁻¹

                                     Mole-CH₄  =  0.25 mol

Moles of He;
                                     Mole-He  =  2 g / 4 g.mol⁻¹

                                     Mole-He  =  0.5 mol

Now adding moles of both gases,
          
                                     =  0.25 + 0.5

                                     =  0.75 mol

Therefore, when,

                          1 mole of gas occupy  =  22.4 L of Volume

Then,

                            0.75 mol will occupy  =  X L of Volume

Solving for X,
                                   X  =  (0.75 mol × 22.4 L) ÷ 1 mole

                                   X  =  16.8 L

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