During combustion, methane yields carbon dioxide and water. the unbalanced equation for this reaction is: ch4(g)+o2(g) → co2(g)+ h2o(l)

what will the mole ratios for the balanced equation be? what coefficients are needed in order to balance the equation?

CH₄(g)  +  2O₂(g) → CO₂(g)  +  2H₂O(l)

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2

Explanation:

CH4(g)+O2(g) → CO2(g)+ H2O(l)

To balance a chemical equation, you must have the same mole of each element in both sides of the reaction (reactant side and product side)

CH₄(g)  +  2O₂(g) → CO₂(g)  +  2H₂O(l)

2 C

8 H

8 O

In  both side.

It takes 1 mole of methane to react with 2 mole of oxygen in order to produce 1 mol of dioxide and 2 mole of water.

Mole ratios for the balanced equation be:

Mole ratios for the balanced equation be:

1:2, 2:1, 1:1, 1:2, 2:2  - We should compare each compound

1 mol methane → 2 mole of oxygen

2 mole of oxygen → 1 mol of methane

2 mole of oxygen → 1 mol of dioxide

1 mole of dioxide → 1 mol of Methane

2 mole of water → 2 mole of oxygen (the same as opposite)

1 mol methane →  2 mole of water

2 mol of water → 1 mol of methane

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole