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Sliva [168]
2 years ago
9

How many grams of soda ash would be needed to produce 1.00 kg of sodium bicarbonate?

Chemistry
1 answer:
GaryK [48]2 years ago
5 0

Answer:

736.1 g of soda ash

Explanation:

Data given:

Amount of sodium bicarbonate = 1 Kg

convert Kg to g

1 Kg = 1000 g

Amount of soda ash needed = ?

Solution:

First we look for the reaction by which soda ash gives sodium bicarbonate.

Chemical formulas

c NaHCO₃

Soda Ash = Na₂CO₃

So the reaction is given below:

                    Na₂CO₃ + H₂O + CO₂ ------------> 2 NaHCO₃

So,

now look for the amount of soda ash and NaHCO₃

                   Na₂CO₃ + H₂O + CO₂ ------------> 2 NaHCO₃

                     1 mole                                           2 mole

As,

one mole of Na₂CO₃ (soda ash) gives 2 moles of sodium bicarbonate (NaHCO₃).

Now convert moles to mass

For which molar masses are given below

Molar mass of Na₂CO₃ = 106 g/mol

Molar mass of NaHCO₃ = 72 g/mol

So,

         Na₂CO₃      +       H₂O     +     CO₂      ------------>    2 NaHCO₃

   1 mole (106 g/mol)                                                    2 mole (72 g/mol)

          106 g                                                                            144 g

Apply unity formula

            106 g of Na₂CO₃ ≅ 144 grams NaHCO₃

                X g of Na₂CO₃ ≅ 1000 grams NaHCO₃

Do cross multiplication

              X g of Na₂CO₃ = 106 g x 1000 g / 144 g

               X g of Na₂CO₃ = 736.1 g

So, 736.1 g of soda ash needed to produce 1000 g or 1 kg of sodium bicarbonate.

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Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide
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<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

          = 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

                            = 6.293 g

Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>
  • We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

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