Question

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

Answer 1

The answer to your question is: 2 grams of methane

Step-by-step 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 = (PV)/(RT)`

`n = ((1.77)(9.15))/((0.082)(330))`

`n = (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