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3 years ago

How do you balance this equation?

Chemistry

1 answer:

anyanavicka [17]3 years ago

5 0

Answer:

HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O

The coefficients are: 1, 1, 1, 1, 1

Explanation:

_HC₂H₃O₂ + _NaHCO₃ —> _NaC₂H₃O₂ + _CO₂ + _H₂O

To balance an equation, we simply do a head count of the individual elements and ensure they are balanced on both side.

For the above equation, we shall balance it as :

HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O

Reactant:

H = 5

C = 3

O = 5

Na = 1

Product:

H = 5

C = 3

O = 5

Na = 1

From the above, we can see that each element is the same on both side of the equation. Thus the equation is already balanced

HC₂H₃O₂ + NaHCO₃ —> NaC₂H₃O₂ + CO₂ + H₂O

The coefficients are: 1, 1, 1, 1, 1

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Answer is: the molarity of the sulfuric acid is 7.14 M.
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3 years ago

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3 years ago

Acrylonitrile is an important building block for synthetic fibers and plastics. Over 2 billion pounds of acrylonitrile are produ

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Answer:

There can be prepared 693.512 kg of acrylonitrile

Explanation:

Propene has molar mass of 42.08g/mole

ammonia has molar mass of 17.03g/ mole

O2 has molar mass of 2*16 = 32g/mole

acrylonitrile has molar mass of 53.06g /mole

Step 1: finding limiting reagent

2C3H6 + 2NH3 + 3O2 → 2C3H3N + 6H2O

We see that the ratio 2:2:3:2:6 is

This means that for 2 mole C3H6 we have 2 mole 2NH3 and 3 mole 3O2, also there will be produced 2 mole C3H3N and 6 mole H2O

Step2 : Calculating moles

moles propene: 550000g / 42.08g/mole = 13070.34 moles

moles ammonia: 650000g/ 17.03g/mole = 38167.94 moles

moles O2: 900000g/ 32g/mole = 28125moles

The limiting reagent is propene:

⇒propene will be consumed completely

⇒ammonia will consume 13070.34 moles ⇒ there will remain 25097.6 moles

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Step 3: Calculating moles of acrylonitrile

There will be formed 13070.34 moles of acrylonitrile

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There can be prepared 693.512 kg of acrylonitrile

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