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hjlf
3 years ago
5

Sodium chloride can be produced from reacting sodium metal with chlorine gas. Carly reacts 2.6 g of sodium with 5.0 g of chlorin

e gas. She recovers 5.8 g of sodium chloride from the reaction. What is her percent yield? Carly's percent yield is ___ %
Chemistry
1 answer:
Artemon [7]3 years ago
7 0

Answer:

87.75%

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Na + Cl₂ —> 2NaCl

Next, we shall determine the masses of Na and Cl₂ that reacted and the mass of NaCl produced from the balanced equation. This can be obtained as follow:

Molar mass of Na = 23 g/mol

Mass of Na from the balanced equation = 2 × 23 = 46 g

Molar mass of Cl₂ = 2 × 35. 5 = 71 g/mol

Mass of Cl₂ from the balanced equation = 1 × 71 = 71 g

Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol

Mass of NaCl from the balanced equation = 2 × 58.5 = 117 g

Summary:

From the balanced equation above,

46 g of Na reacted with 71 g of Cl₂ to produce 117 g of NaCl.

Next, we shall determine the limiting reactant.

This can be obtained as follow:

From the balanced equation above,

46 g of Na reacted with 71 g of Cl₂.

Therefore, 2.6 g of Na will react with = (2.6 × 71)/46 = 4.01 g of Cl₂.

From the calculations made above, we can see that only 4.01 g of Cl₂ at of 5 g given in question reacted completely with 2.6 g of Na. Therefore, Na is the limiting reactant and Cl₂ is the excess reactant.

Next, we shall determine the theoretical yield of NaCl. The limiting reactant will be used to obtain the theoretical yield since all of it is consumed in the reaction.

The limiting reactant is Na and the theoretical yield of NaCl can be obtained as follow:

From the balanced equation above,

46 g of Na reacted to produce 117 g of NaCl.

Therefore, 2.6 g of Na will react to produce = (2.6 × 117)/46 = 6.61 g of NaCl.

Thus the theoretical yield of NaCl is 6.61 g

Finally, we shall determine the percentage yield of NaCl. This can be obtained as follow:

Actual yield of NaCl = 5.8 g

Theoretical yield of NaCl = 6.61 g

Percentage yield =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield = 5.8/6.61 ×100

Percentage yield of NaCl = 87.75%

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How many moles of potassium nitrate are required to make 550 mL of a 2.1M solution?
stira [4]

Answer:

1.155 moles of potassium nitrate are required to make 550 mL of a 2.1M solution.

Explanation:

In a mixture, the chemical present in the greatest amount is called a solvent, while the other components are called solutes.

Molarity is a unit of concentration of a solution and indicates the amount of moles of solute that appear dissolved in each liter of the mixture. In other words, the Molarity (M) or Molar Concentration is the number of moles of solute that are dissolved in a given volume.

The Molarity of a solution is determined by the following expression:

Molatity( M)=\frac{number of moles of solute}{Volume}

Molarity is expressed in units (\frac{moles}{liter}).

In this case:

  • Molarity= 2.1 M
  • number of moles of solute= ?
  • Volume= 550 mL= 0.550 L (being 1L=1000 mL)

Replacing:

2.1 M= 2.1 \frac{moles}{liter} =\frac{number of moles of solute}{0.550 liters}

Solving:

number of moles of solute= 2.1 M* 0.550 L

number of moles of solute= 1.155 moles

1.155 moles of potassium nitrate are required to make 550 mL of a 2.1M solution.

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