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Elanso [62]
3 years ago
10

Equation balancing

Chemistry
1 answer:
meriva3 years ago
4 0

<u>Answer:</u>

<u>For a:</u> The balanced equation is 2S(s)+3O_2(g)\rightarrow 2SO_3(g)

<u>For c:</u> The balanced equation is 2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

<u>Explanation:</u>

A balanced chemical equation is one where all the individual atoms are equal on both sides of the reaction. It follows the law of conservation of mass.

  • <u>For (a):</u>

The given unbalanced equation follows:

S(s)+O_2(g)\rightarrow SO_3(g)

To balance the equation, we must balance the atoms by adding 2 infront of both S(s) and SO_3 and 3 in front of O_2

For the balanced chemical equation:

2S(s)+3O_2(g)\rightarrow 2SO_3(g)

  • <u>For (b):</u>

The given balanced equation follows:

2Al(s)+3Cl_2(g)\rightarrow 2AlCl_3(s)

The given equation is already balanced.

  • <u>For (c):</u>

The given unbalanced equation follows:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+H_2O(l)

To balance the equation, we must balance the atoms by adding 2 infront of H_2O(l)

For the balanced chemical equation:

2NaOH(s)+H_2SO_4(aq)\rightarrow Na_2SO_4(aq)+2H_2O(l)

  • <u>For (d):</u>

The given balanced equation follows:

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(g)

The given equation is already balanced.

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UGRENT! Please help showing all work
agasfer [191]

Answer:

a. The limiting reactant is Ca(OH)₂

b. The theoretical yield of CaCl₂ is approximately 621.488 grams

c. The percentage yield of CaCl₂ is approximately 47.06%

Explanation:

a. The given chemical reaction is presented as follows;

Ca(OH)₂ + 2HCl → CaCl₂ + 2H₂O

Therefore;

One mole of Ca(OH)₂ reacts with two moles of HCl to produce one mole of CaCl₂ and two moles of water H₂O

The mass of HCl in an experiment, m₁ = 229.70 g

The mass of Ca(OH)₂ in an experiment, m₂ = 207.48 g

The molar mass of HCl, MM₁ = 36.458 g/mol

The molar mass of Ca(OH)₂, MM₂ =74.093 g/mol

The number of moles of HCl, present, n₁ = m₁/MM₁

∴ n₁ = m₁/MM₁ = 229.70 g/(36.458 g/mol) ≈ 6.3 moles

The number of moles of Ca(OH)₂, present, n₂ = m₂/MM₂

∴ n₂ = m₂/MM₂ = 207.48 g/(74.093 g/mol) ≈ 2.8 moles

The number of moles of Ca(OH)₂, present, n₂ = 2.8 moles

According the chemical reaction equation the number of moles of HCl the 2.8 moles of Ca(OH)₂ will react with, = 2.8 × 2 moles = 5.6 moles of HCl

Therefore, there is an excess HCl in the reaction and Ca(OH)₂ is the limiting reactant

b. According the chemical reaction equation the number of moles of CaCl₂ produced in he reaction by the 2.8 moles of Ca(OH)₂ = 2.8 × 2 moles = 5.6 moles of CaCl₂

The molar mass of CaCl₂ = 110.98 g/mol

The mass of the 5.6 moles of CaCl₂ = 5.6 moles × 110.98 g/mol ≈ 621.488 grams

The theoretical yield of CaCl₂ ≈ 621.488 grams

c. Given that the actual mass of CaCl₂ produced = 292.5 grams, we have;

The percentage yield of CaCl₂ = The actual yield/(The theoretical yield) × 100

∴ The percentage yield of CaCl₂ = (292.5 g)/(621.488 g) × 100 ≈ 47.0644646397%

The percentage yield of CaCl₂ ≈ 47.06%.

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