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NARA [144]
3 years ago
6

Write a balanced chemical equation for the fermentation of sucrose (C12H22O11) by yeasts in which the aqueous sugar reacts with

water to form aqueous ethyl alcohol (C2H5OH) and carbon dioxide gas.
Chemistry
1 answer:
bogdanovich [222]3 years ago
4 0

Answer:

C_{12}H_{22}O_{11}(aq)+H_{2}O(aq)=4C_{2}H_{5}OH(aq)+4CO_{2}(g)

Explanation:

1. First write the formula of each compound:

Sucrose : C_{12}H_{22}O_{11}

Water : H_{2}O

Ethyl alcohol : C_{2}H_{5}OH

Carbon dioxide : CO_{2}

2. Write the equation with the reactants on the left and the products on the right:

C_{12}H_{22}O_{11}+H_{2}O=C_{2}H_{5}OH+CO_{2}

3. Put the states of matter of each compound in the reaction, as the problem says:

C_{12}H_{22}O_{11}(aq)+H_{2}O(aq)=C_{2}H_{5}OH(aq)+CO_{2}(g)

aq = aqueous solution

g =  gas

4. Balance each element.

The number of each atom must be the same on the left and on the right.

C_{12}H_{22}O_{11}(aq)+H_{2}O(aq)=4C_{2}H_{5}OH(aq)+4CO_{2}(g)

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The compound known as diethyl ether, commonly referred to as ether, contains carbon, hydrogen, and oxygen. A 1.376 g sample of e
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Answer:

The answer to your question is: C₄H₁₀O

Explanation:

Data

          CxHyOz

mass sample : 1.376 g

mass CO₂ = 3.268 g

mass H₂O = 1.672 g

Process

Reaction

                      CxHyOz  + O₂ ⇒   CO₂  +  H₂O

1.- Calculate the moles and mass of carbon

Molecular mass CO₂ = 44g

                      44 g of CO₂ --------------  12 g of C

                      3.268 g of CO₂  --------    x

                         x = (3.268 x 12) / 44

                        x = 0.891 g of Carbon

                       12 g of carbon -----------  1 mol

                       0.891 g of C     ----------   x

                       x = (0.891 x 1) / 12

                       x = 0.0743 moles of carbon

2.- Calculate the moles and mass of hydrogen

                      18 g of water --------------- 2 g of H

                      1.672 g of H₂O ------------  x

                      x = (1.672 x 2) / 18

                      x = 0.186 g of hydrogen

                      1 g of hydrogen ------------  1 mol of H

                      0.186 g of H       ------------  x

                      x = (0.186 x 1) / 1

                      x = 0.186 moles of H

3.- Calculate the mass of Oxygen and its moles

Mass of Oxygen = 1.376 - 0.891 - 0.186

                           = 0.299 g of O₂

Moles of Oxygen

                             16 g of Oxygen ---------------- 1 mol

                             0.299 g of O    -----------------  x

                             x = (0.299 x 1) / 16

                             x = 0.019 moles of Oxygen

4.- Divide by the lowest number of moles

Carbon         0.0743/ 0.019 = 3.9 ≈ 4.0

Hydrogen     0.186/ 0.019 = 9.7 = 10

Oxygen         0.019/ 0.019 = 1

5.- Write the empirical formula

                              C₄H₁₀O                  

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