Gaseous methane (CH4) reacts with gaseous oxygen gas (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). If 28.2 g of carbon dioxide is produced from the reaction of 15.1 g of methane and 81.2 g of oxygen gas, Calculate the percent yield of carbon dioxide. Be sure your answer has the correct number of significant digits in it.

68.1% is percent yield of the reaction

Explanation:

The reaction of methane with oxygen is:

CH₄ + 2O₂ → CO₂ + 2H₂O

Where 2 moles of oxygen react per mole of CH₄

Percent yield is:

Actual yield (28.2g CO₂) / Theoretical yield * 100

To solve this question we need to find theoretical yield finding limiting reactant :

Moles CH₄:

15.1g CH₄ * (1mol / 16.04g) = 0.9414 moles

Moles O₂:

81.2g * (1mol / 32g) = 2.54 moles

For a complete reaction of 0.9414 moles of CH₄ are needed:

0.9414 moles CH₄ * (2 mol O₂ / 1mol CH₄) = 1.88 moles of O₂. As there are 2.54 moles, O₂ is in excess and CH₄ is limiting reactant

In theoretical yield, the moles of methane added = Moles of CO₂ produced. That is 0.9414 moles CO₂. In grams = Theoretical yield:

0.9414 moles CO₂ * (44.01g / mol) = 41.43g CO₂

Percent yield: 28.2g CO₂ / 41.43g CO₂ * 100=

beta decay is the answer

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.