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3 years ago

find the percent yield,if 6.0g of Licl are actually produced and the theoretical yield is 35.4g of Licl.

Chemistry

1 answer:

drek231 [11]3 years ago

8 0

Answer:

Percent yield = 17%

Explanation:

Given data:

Actual yield of lithium chloride = 6.0 g

Theoretical yield of lithium chloride = 35.4 g

Percent yield = ?

Solution:

Formula:

Percent yield = (actual yield / theoretical yield )× 100

Now we will put the values in formula.

Percent yield = (6.0 g/ 35.4 g)× 100

Percent yield = 0.17 × 100

Percent yield = 17%

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The number of mole of ammonium ion, NH₄⁺ in the solution is 0.175 mole

We'll begin by calculating the number of mole of NH₄NO₃ in the solution. This can be obtained as follow:

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The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.

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Answer:

a) The theoretical yield is 408.45g of $BaSO_{4}$

b) Percent yield = $\frac{realyield}{408.45g}*100$

Explanation:

1. First determine the numer of moles of $BaCl_{2}$ and $Na_{2}SO_{4}$ .

Molarity is expressed as:

M= $\frac{molessolute}{Lsolution}$

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M= $\frac{1.75molesBaCl_{2}}{1Lsolution}$

Therefore there are 1.75 moles of $BaCl_{2}$

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M= $\frac{2.0moles[tex]Na_{2}SO_{4}$ }{1Lsolution}[/tex]

Therefore there are 2.0 moles of $Na_{2}SO_{4}$

2. Write the balanced chemical equation for the synthesis of the barium white pigment, $BaSO_{4}$ :

$BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl$

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the $BaCl_{2}$ :

$\frac{1.75}{1}=1.75$

- For the $Na_{2}SO_{4}$ :

$\frac{2.0}{1}=2.0$

As the $BaCl_{2}$ is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

$1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}$

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = $\frac{realyield}{theoreticalyield}*100$

Percent yield = $\frac{realyield}{408.45g}*100$

The real yield is the quantity of barium white pigment you obtained in the laboratory.

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Answer:

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find the percent yield,if 6.0g of Licl are actually produced and the theoretical yield is 35.4g of Licl.​

find the percent yield,if 6.0g of Licl are actually produced and the theoretical yield is 35.4g of Licl.