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-BARSIC- [3]
2 years ago
5

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?
Chemistry
1 answer:
Korolek [52]2 years ago
8 0

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{C_6H_{12}O_6} \longrightarrow 2 \ C_2H_6O + 2 \ CO_2

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

<u>Method 1: using the </u><u>mass of glucose</u>

Mr of glucose

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

= 180

Moles of glucose reacted

= 200 ÷180

= \frac{10}{9} mol

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

Amount of ethanol

= 2(\frac{10}{9} )

= \frac{20}{9} mol

Mass of ethanol

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

= \frac{20}{9}(46)

= 102 g (3 s.f.)

<u>Method 2: using </u><u>mass of carbon dioxide</u><u> produced</u>

Mole of carbon dioxide produced

= 97.7 ÷[12 +2(16)]

= 97.7 ÷44

= \frac{977}{440} mol

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

Moles of ethanol formed= \frac{977}{440} mol

Mass of ethanol formed

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

= 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!

  • brainly.com/question/1685725
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A radioactive material, with half-life of six months, has 100 thousand unstable nuclei.
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See Explanation

Explanation:

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Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;

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3 0
3 years ago
A buffer solution is made that is 0.347 M in H2C2O4 and 0.347 M KHC2O4.
irga5000 [103]

Answer:

1. pH = 1.23.

2. H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Explanation:

Hello!

1. In this case, for the ionization of H2C2O4, we can write:

H_2C_2O_4\rightleftharpoons HC_2O_4^-+H^+

It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:

pH=pKa+log(\frac{[base]}{[acid]} )

Whereas the pKa is:

pKa=-log(Ka)=-log(5.90x10^{-2})=1.23

The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:

pH=1.23+log(\frac{0.347M}{0.347M} )\\\\pH=1.23+0\\\\pH=1.23

2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:

n_{acid}=0.347mol/L*1.00L=0.347mol\\\\n_{acid}^{remaining}=0.347mol-0.070mol=0.277mol

It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:

H_2C_2O_4(aq) +OH^-(aq)\rightarrow HC_2O_4^-(aq)+H_2O(l)

Which is also shown in net ionic notation.

Best regards!

4 0
3 years ago
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