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valina [46]
3 years ago
13

Balancing Chemical Equations Worksheet

Chemistry
1 answer:
Goshia [24]3 years ago
5 0

The balanced chemical equation of the reactions given is as follows:

  • 2LiHCO3 -----> Li2CO3 + H2O + CO2
  • 2 N2 + 5 O2 -----> 2 N2O5
  • MgBr2 + KOH -----> KBr + Mg(OH)2
  • Mn + 2 CuCl -----> 2 Cu + MnCl2
  • 8 Zn + S8 -----> 8 ZnS
  • 2 NaOH + H2SO4 -----> 2 H2O + Na2SO4
  • 2 K + 2 H2O -----> 2 KOH + H2
  • C5H12 + 8 O2 -----> 6H2O + 5 CO2
  • 2 KOH + H2CO3 -----> 2 H2O + K2CO3
  • C4H802 + 6 O2 -----> 4 H20 + 4 CO2
  • 16 Al + 3 S8 ---> 8 Al2S3

<h3>How to balance chemical equations</h3>

Balancing of chemical equations is the process of adding numerical coefficients in front of moles of reactants and products to ensure that the moles of atoms of elements of the reactants is equal to the moles of atoms of products formed.

The balanced chemical equation of the reactions given is as follows:

  • 2LiHCO3 -----> Li2CO3 + H2O + CO2
  • 2 N2 + 5 O2 -----> 2 N2O5
  • MgBr2 + KOH -----> KBr + Mg(OH)2
  • Mn + 2 CuCl -----> 2 Cu + MnCl2
  • 8 Zn + S8 -----> 8 ZnS
  • 2 NaOH + H2SO4 -----> 2 H2O + Na2SO4
  • 2 K + 2 H2O -----> 2 KOH + H2
  • C5H12 + 8 O2 -----> 6H2O + 5 CO2
  • 2 KOH + H2CO3 -----> 2 H2O + K2CO3
  • C4H802 + 6 O2 -----> 4 H20 + 4 CO2
  • 16 Al + 3 S8 ---> 8 Al2S3

Learn more about balancing of chemical equations at: brainly.com/question/15428811

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Pb(SO4)2 + 4 LiNO3 → Pb(NO3)4 + 2 Li2SO4
Anvisha [2.4K]

Answer:

4.5 moles of lithium sulfate are produced.

Explanation:

Given data:

Number of moles of lead sulfate = 2.25 mol

Number of moles of lithium nitrate = 9.62 mol

Number of moles of lithium sulfate = ?

Solution:

Chemical equation:

Pb(SO₄)₂ + 4LiNO₃      →     Pb(NO₃)₄ + 2Li₂SO₄

Now we will compare the moles of lithium sulfate with lead sulfate and lithium nitrate.

                       Pb(SO₄)₂        :         Li₂SO₄

                            1                :             2

                          2.25           :          2/1×2.25 = 4.5 mol

                       LiNO₃            :             Li₂SO₄

                           4                :                2

                           9.62           :             2/4×9.62 = 4.81 mol

Pb(SO₄)₂  produces less number of moles of Li₂SO₄ thus it will act as limiting reactant and limit the yield of  Li₂SO₄.      

7 0
3 years ago
Ammonia and oxygen react to form nitrogen and water.
Nata [24]

Answer:

A. 19.2 g of O2.

B. 3.79 g of N2.

C. 54 g of H2O.

Explanation:

The balanced equation for the reaction is given below:

4NH3(g) + 3O2(g) → 2N2+ 6H2O(g)

Next, we shall determine the masses of NH3 and O2 that reacted and the masses of N2 and H2O produced from the balanced equation.

This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 3 x 32 = 96 g

Molar mass of N2 = 2x14 = 28 g/mol

Mass of N2 from the balanced equation = 2 x 28 = 56 g

Molar mass of H2O = (2x1) + 16 = 18 g/mol

Mass of H2O from the balanced equation = 6 x 18 = 108 g

Summary:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2 to produce 56 g of N2 and 108 g of H2O.

A. Determination of the mass of O2 needed to react with 13.6 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted with 96 g of O2.

Therefore, 13.6 g of NH3 will react with = (13.6 x 96)/68 = 19.2 g of O2.

Therefore, 19.2 g of O2 are needed for the reaction.

B. Determination of the mass of N2 produced when 6.50 g of O2 react.

This is illustrated below:

From the balanced equation above,

96 g of O2 reacted to produce 56 g of N2.

Therefore, 6.5 g of O2 will react to produce = (6.5 x 56)/96 = 3.79 g of N2.

Therefore, 3.79 g of N2 were produced from the reaction.

C. Determination of the mass of H2O formed from the reaction of 34 g of NH3.

This is illustrated below:

From the balanced equation above,

68 g of NH3 reacted to 108 g of H2O.

Therefore, 34 g of NH3 will react to produce = (34 x 108)/68 = 54 g of H2O.

Therefore, 54 g of H2O were obtained from the reaction.

4 0
3 years ago
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