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Gekata [30.6K]
2 years ago
12

Help would be majorly appreciated

Chemistry
1 answer:
Sophie [7]2 years ago
6 0

Answer:

a) the atomic number is 15

b) the mass number is 15+16 = 31

c) element is phosphorus

d)Group 15 period 3

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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

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

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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

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<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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