CH4 + 202 → CO2 + 2H2O
How many moles O2 needed to produce 4 moles of H2O?

4 mol O₂

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

[RxN - Balanced]   CH₄ + 2O₂ → CO₂ + 2H₂O

[Given]   4 mol H₂O

[Solve]   x mol O₂

Step 2: Identify Conversions

[RxN] 2 mol H₂O → 2 mol O₂

Step 3: Stoichiometry

Moles of O₂= 4

Given

Reaction

CH₄ + 20₂ → CO₂ + 2H₂O

4 moles of H₂O

Required

Moles of O₂

Solution

In a chemical equation, the reaction coefficient shows the mole ratio of the compounds involved in the reaction, reactants or products

From the equation, mol ratio of H₂O  : O₂ = 2 : 2, so moles O₂ :

= 2/2 x moles H₂O

= 2/2 x 24

= 4 moles

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

hi my dude or dudette