Question

Correct first answer= Brianliest!PLEASE WOULD REALLY APPRECIATE IF YOU’D HELP ME OUT:’(

Answer 1

Answer:

The answer to your question is: 2 grams of methane

Explanation:

Data

V = 9.15 l

P = 1.77 atm

T = 57° C = 330 °K

Ar mass = 19 g

CH4 mass = ?

Formula

PV = nRT

Process

                     $n = \frac{PV}{RT}$

                      $n = \frac{(1.77)(9.15)}{(0.082)(330)}$

                      $n = \frac{16.26}{27.06}$

                             n = 0.6

Argon

                     40 g of Ar -------------------- 1 mol

                      19 g         ---------------------   x

                                 x = 0.475 mol of Ar

moles of CH4 = 0.6 - 0.475

                       = 0.125

                     

Methane

                            16 g of CH4 ---------------  1mol

                             x                 ----------------  0.125 mol

                            x = 2 grams