Yeast converts glucose to ethanol and carbon dioxide during a process known as anaerobic fermentation. The chemical reaction is C6H12O6 --> 2 C2H6O + 2 CO2 if 200.0 g of glucose are used and 97.7 g of CO2 are produced in a reaction that goes to completion, how many g or ethanol (C2H6O) are formed?

Respuesta :

Answer:

102g

Explanation:

To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.

In this case, the given equation is already balanced.

[tex]\tex{C_6H_{12}O_6} \longrightarrow 2 \ C_2H_6O + 2 \ CO_2[/tex]

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.

Mole= Mass ÷Mr

Mass= Mole ×Mr

Method 1: using the mass of glucose

Mr of glucose

= 6(12) +12(1) +6(16)

= 180

Moles of glucose reacted

= 200 ÷180

= [tex]\frac{10}{9}[/tex] mol

Amount of ethanol formed: moles of glucose reacted= 2: 1

Amount of ethanol

= [tex]2(\frac{10}{9} )[/tex]

= [tex]\frac{20}{9}[/tex] mol

Mass of ethanol

= [tex]\frac{20}{9} \times[2(12)+6+16][/tex]

= [tex]\frac{20}{9}(46)[/tex]

= 102 g (3 s.f.)

Method 2: using mass of carbon dioxide produced

Mole of carbon dioxide produced

= 97.7 ÷[12 +2(16)]

= 97.7 ÷44

= [tex]\frac{977}{440}[/tex] mol

Moles of ethanol: moles of carbon dioxide= 1: 1

Moles of ethanol formed= [tex]\frac{977}{440}[/tex] mol

Mass of ethanol formed

= [tex]\frac{977}{440} \times[2(12)+6+16][/tex]

= 102 g (3 s.f.)

Thus, 102 g of ethanol are formed.

Additional:

For a similar question on mass and mole ratio, do check out the following!

  • https://brainly.com/question/1685725