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Leona [35]
3 years ago
7

How to balance equations

Chemistry
1 answer:
dezoksy [38]3 years ago
4 0

start the balancing by writing down how many atoms there are per element. we’ll use this as an example:

C3H8 + O2 --> H2O + CO2

C = 3 C = 1

H = 8 H = 2

O = 2 O = 3

balance the carbon first, as it is easiest to do. add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon atoms on the left of the equation:

C3H8 + O2 --> H2O + (3)CO2

now there are 3 carbon atoms on each side. however, when you do this, you multiply the amount of oxygen atoms you had. therefore, now, there are 6 carbon atoms in 3CO2, plus that other oxygen atom in H2O. you now have 7 O atoms instead of 3.

C = 3 C = 3

H = 8 H = 2

O = 2 O = 7

now let’s move on to the hydrogen atoms.

C3H8 + O2 --> H2O + 3CO2

you have 8 hydrogen atoms on the left side, and 2 on the right. in order to balance them, you have to multiply the right side’s hydrogen atoms by 4. 4(2) = 8.

C3H8 + O2 --> (4)H2O + 3CO2

now both hydrogen and carbon atoms are balanced. same amount on both sides. however, your oxygen atoms have changed due to the multiplying (right side). you now have 10 of them.

C = 3 C = 3

H = 8 H = 8

O = 2 O = 10

now we balance the oxygen atoms. multiply the left side of the equation’s oxygen atoms by 5. 5(2) = 10

C3H8 + (5)O2 --> 4H2O + 3CO2

the chemical equation is all balanced. basically, just multiply with numbers until it equals the same amount on both sides.

C = 3 C = 3

H = 8 H = 8

O = 10 O = 10

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Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Cu: 3 moles
  • HNO₃: 8 moles
  • Cu(NO₃)₂: 3 mole
  • NO: 2 moles
  • H₂O: 4 moles

The molar mass of the compounds is:

  • Cu: 63.55 g/mole
  • HNO₃: 63 g/mole
  • Cu(NO₃)₂: 187.55 g/mole
  • NO: 30 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Cu: 3 moles ×63.55 g/mole= 190.65 grams
  • HNO₃: 8 moles ×63 g/mole= 504 grams
  • Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
  • NO: 2 moles ×30 g/mole= 60 grams
  • H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3}  x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }

<u><em>mass of Cu(NO₃)₂=  5.2634 grams</em></u>

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

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