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Bas_tet [7]
3 years ago
8

Suppose 200.0 mL of a 2.50 M solution of sodium hydroxide is combined with 400.0 mL of a 1.50 M solution of iron(III) nitrate. W

hat is the theoretical yield of precipitate? If 14.8 g of the precipitate actually formed, what is the percent yield?

Chemistry
1 answer:
-BARSIC- [3]3 years ago
6 0

Answer: 83%

Explanation:

The detailed solution is shown in the image attached. First we must work out the balanced reaction equation because accurate solution of the problem must be based on the stoichiometry of the reaction. From the given concentration and volume of reactants, we calculate the amount of substance reacted hence identify the limiting reactant. Lastly we use simple proportion to obtain the theoretical yield of the precipitate. This is now used to calculate the actual yield as shown in the solution attached.

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Agriculture was practiced for thousands of years without the use of artificial chemicals. Artificial fertilizers were first created during the mid-19th century. These new agricultural techniques, while beneficial in the short term, had an Institute of Plant Industry to improve traditional farming methods in India.

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Which of the following measurements is expressed to three significant figures? 7.30 × 10–7 km 0.007 m 7077 mg 0.070 mm
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7.30 x 10^-7 km. the others have 4 significant figures.

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Explain how the mass of a single atom in an element (in amu), the mass of one mole of that element, and Avogadro’s number are re
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Answer: For 1 mole of a single atom it is equal to its molar mass. And a single atom, 1 mole is equal to the Avogadro's Number.

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4 0
3 years ago
agcl molar masA 250.0 g sample of a white solid is known to be a mixture of KNO3, BaCl2, and NaCl. When 100.0 g of this mixture
tatiyna

Answer:

a. BaSO₄ and AgCl.

b. 150.0g of BaCl₂, 50.0g of NaCl and 50.0g of KNO₃

Explanation:

Barium, Ba, from BaCl₂ reacts with the SO₄²⁻ of H₂SO₄ to produce BaSO₄, an insoluble white salt.

The reaction is:

BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl

Also, Chlorides from BaCl₂ (2Cl⁻) and NaCl (1Cl⁻) react with AgNO₃ to produce AgCl, another white insoluble salt, thus:

Cl⁻ + AgNO₃ → AgCl + NO₃⁻

a. Thus, formulas of the two precipitates are: BaSO₄ and AgCl

b. Moles of BaSO₄ in 67.3g (Molar mass BaSO₄: 233.38g/mol) are:

67.3g × (1 mol / 233.38g) = 0.2884 moles of BaSO₄ = moles of BaCl₂ <em>Because 1 mole of BaCl₂ produces 1 mole of BaSO₄</em>

Now, as molar mass of BaCl₂ is 208.23g/mol, the mass of BaCl₂ in the mixture of 100.0g is:

0.2884 moles of BaCl₂ ₓ (208.23g /mol) = 60.0g of BaCl₂ in 100g of the mixture

Moles of the AgCl produced (Molar mass AgCl: 143.32g/mol) are:

197.96g ₓ (1mol / 143.32g) = 1.38 moles of AgCl.

As moles of Cl⁻ that comes from BaCl₂ are 0.2884 moles×2×1.5 (1.5 because the sample is 150.0g not 100.0g as in the initial reaction)

= 0.8652 moles of BaCl₂, that means moles of NaCl are:

1.38mol - 0.8652mol = 0.5148 moles of NaCl (Molar mass 58.44g/mol):

Mass NaCl in 150g =

0.5148mol NaCl × (58.44g/mol) = <em>30.0g of NaCl in 150.0g</em>

<em></em>

That means, in the 250.0g of sample, the mass of BaCl₂ is:

60.0g BaCl₂ ₓ (250.0g / 100g) = <em>150.0g of BaCl₂</em>

Mass of NaCl is:

30.0g NaCl ₓ (250.0g / 150g) =<em> 50.0g of NaCl</em>

<em></em>

As the total mass of the mixture is 250.0g, the another 50.0g must come from KNO₃, thus, there are <em>50.0g of KNO₃.</em>

4 0
3 years ago
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