Chemistry Help!!!!! If 6.0g of CH4 and 5.0g of O2 are used, what is the mass of CO2 produced?

Molar mass of CH₄ = (12.0 + 1.0×4) g/mol = 16.0 g/mol
Initial number of moles of CH₄ used = (6.0 g) / (16.0 g/mol) = 0.375 mol

Molar mass of O₂ = 16.0 × 2 = 32.0 g/mol
Initial number of moles of O₂ = (5.0 g) / (32.0 g) = 0.1563 mol

Equation for the reaction:
CH₄ + 2O₂ → CO₂ + 2H₂O

According to the above equation, mole ratio CH₄ : O₂ = 1 : 2
When O₂ completely reacts, CH₄ needed = (0.1563 mol) × (1/2) = 0.0782 mol < 0.375 mol
Hence, CH₄ is in excess, and O₂ completely reacts.

According to the above equation, mole ratio O₂ : CO₂ = 2 : 1
Number of moles of O₂ reacted = 5/32.0 mol
Number of moles of CO₂ produced = (0.1563 mol) × (1/2) = 0.0782 mol

Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
Mass of CO₂ produced = (0.0782 mol) × (44.0 g/mol) = 3.44 g

Balanced equation:

CH4 + 2 O2 --> CO2 + 2 H2O

Convert each mass to moles:
6.0 g CH4 / 16.0 g/mol = 0.375 mol CH4
5.0 g O2 / 32.0 g/mol = 0.156 mol O2

This is a limiting reactant type of problem. Using the balanced equation, in order to consume all of the CH4 available, you would need:
0.375 mol CH4 X (2 mol O2 / 1 mol CH4) = 0.75 mol O2
Because you only have 0.156 mol of O2 available, O2 is the limiting reactant. Calculate the mass of CO2 formed based on the amount of O2 available:

0.156 mol O2 X (1 mol CO2 / 2 mol O2) X 44.0 g/mol = 3.4 g CO2 formed

The first step is to write the balanced equation for this reaction.

CH4 + O2 → CO2 + H2O
To balance the H’s, make 2 H2O.
CH4 + O2 → CO2 + 2 H2O
To balance the O’s make O2.
CH4 + O2 → CO2 + 2 H2O
CH4 + 2 O2 → CO2 + 2 H2O

According to the coefficients in this equation, one mole of CH4 reacts with two moles of O2 to produce one mole of CO2 and two moles of H2O. To determine the number of moles, divide the mass the mass of one mole.

For CH4 and O2, ratio = 1:2

CH4 = 12 + 4 = 16 grams
n = 6 ÷ 16 = 0.375
O2 = 32 grams
n = 5 ÷ 32 = 0.15625

Since this is less than twice the number of twice the number of moles of the methane, the oxygen is the limiting reactant. This means all of the oxygen will react. The ratio of moles of O2 to CO2 is 2:1.

For CO2, n = ½ * 0.15625 = 0.078125
Mass of one mole = 12 + 32 = 44 grams
Mass = 0.078125 * 44 = 3.4375 grams

I hope this is helpful for you.

first you need to determine WHICH stoichiometric equation you are going to use and how details the products are from the reactants, we'll go ideal and say its pure O2 with no S2 or N2 present. so ...CH4+ O2----> CO2 + H2O balance this equation results in 1CH4+ 2O2----> 1CO2 + 2H2O ....so now you know from a stoichiometric perspective you will require 2 moles O2 for every 1 mole of CH4. now what are the MWT of CH4(16) and O2(32) so you have 0,375 moles CH4 and 0,15625 moles O2, what does this tell you?...can the reaction proceed?....0.078125......times the MWT of CO2....44.....will tell you the mass.....