Methane burns in the presence of oxygen to produce carbon dioxide and water in the following reaction:

CH4 (s) + 2 O2 (g) --> CO2 (g) + 2 H2O (l)

What mass of methane (in grams) will require 35.15 L of oxygen to fully react? The pressure is 2.35 atm and the temperature is 15 °C. (R = 0.0821 L atm/mol K)

The mass of methane that will require 35.15 L of oxygen to fully react is 27.04 grams

Explanation:

Here we have 2 moles of oxygen gas combining with 1 mole of methane gas

At 2.35 atm  and 15 °C, the volume of oxygen is 35.15 L

Number of moles, n of oxygen is given by the following relation;

Therefore

Therefore, since 2 moles of oxygen gas combining with 1 mole of methane the number of moles of methane combining with 3.375 moles of oxygen = 3.375/2 moles or 1.69 moles of methane

Molar mass of methane = 16.04 g/mol

Therefore, mass of 1.69 moles of methane = 16.04 g/mol × 1.69 mole = 27.064 g

Which means that the mass of methane that will require 35.15 L of oxygen to fully react = 27.04 grams.

i would say b

explanation:

but i think i would need more information than that