Question

What mass of o2 is required to produce 14.5 g of co2 if the reaction has a 65.0% yield? Ch4(g) + 2o2(g) \longrightarrow ⟶ co2(g) + 2h2o(g)?

Answer 1

the balanced equation for the combustion of methane is as follows

CH₄ + 2O₂ --> CO₂ + 2H₂O

stoichiometry of O₂ to CO₂ is 2:1

the percentage yield of CO₂ is 65.0 %

percentage yield = (actual yield / theoretical yield) x 100 %

substituting these values to find the theoretical yield

65 % = (14.5 g / theoretical yield) x 100 %

theoretical yield = 14.5 / 0.65 = 22.3 g

the theoretical yield of CO₂ is 22.3 g

therefore number of moles of CO₂ that was expected to be formed - 22.3 g / 44 g/mol = 0.507 mol

according to molar ratio O₂ to CO₂ ratio is 2:1

therefore number of O₂ moles required to react - 0.507 x2 = 1.01 mol

mass of O₂ required - 1.01 mol x 32 g/mol = 32.32 g

therefore mass of O₂ needed is 32.32 g

Answer 2

The mass of oxygen required to produce 14.5 g of carbon dioxide is 32.32 g.

What is mass?

In chemistry, mass is a property that shows the quantity of the matter.

Mass is calculated in grams and kilograms.

Given, Yield is 65.0%

To produce is 14.5 g of carbon dioxide

The balanced equation is

$\bold{Ch_4(g) + 2O2_(g) \longrightarrow CO_2(g) + 2H_2O(g)}$

Step 1: To calculate the theoretical yield

$\bold{Percentage\; yield = \dfrac{actual\; yield}{theoretical\; yield} \times 100}$

$\bold{65 \% = \dfrac{14.5\;g}{theoretical\; yield} \times 100}$

$\bold{Theoretical\; yield = \dfrac{14.5\;g}{0.65 } =22.3 g}$

The theoretical yield of CO₂ is 22.3 g

Step 2: calculate the moles of CO₂

$\bold{Moles = \dfrac{22.3 g}{44 g/mol } =0.507 mol}$

Step 3: calculate the mass of oxygen

The molar ratio, O₂ to CO₂ ratio is 2:1

Therefore, number of moles of oxygen required is

0.507 x2 = 1.01 mol

Mass of oxygen -volume x density

1.01 mol x 32 g/mol = 32.32 g

Thus, the mass of oxygen required is 32.32 g.

Learn more about mass, here:

https://brainly.com/question/15959704