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3 years ago

Barium nitrate and sodium sulfate solutions can be used to precipitate barium sulfate. How many grams

Chemistry

1 answer:

-Dominant- [34]3 years ago

7 0

Answer:

40.8g of sodium sulfate must be added

Explanation:

The reaction of barium nitrate, Ba(NO₃)₂ with sodium sulfate, Na₂SO₄ is:

Ba(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + BaSO₄(s)

That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:

<em>Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:</em>

75g * (1mol / 261.3g) = 0.287 moles = Moles Na₂SO₄

<em>Mass Na₂SO₄ -Molar mass: 142.04g/mol-:</em>

0.287 moles * (142.04g / mol) =

<h3>40.8g of sodium sulfate must be added</h3>

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CaO + H2O -> Ca(OH)2

yawa3891 [41]

The % yield of Ca(OH)₂ : 62.98%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

CaO + H₂O ⇒ Ca(OH)₂

mass CaO= 4.2 g

mol CaO(MW=56,0774 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{4.2}{56,0774 g/mol}\\\\mol=0.075$

mol Ca(OH)₂ based on mol CaO

mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075

mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical

$\tt mass=mol\times MW\\\\mass=0.075\times 74,093 g/mol\\\\mass=5.557~g$

% yield :

$\tt =\dfrac{actual}{theoretical}\times 100\%\\\\=\dfrac{3.5}{5.557}\times 100\%\\\\=62.98\%$

8 0

2 years ago

A 0.1510 gram sample of a hydrocarbon produces 0.5008 gram CO2 and 0.1282 gram H2O in combustion analysis. Its

Over [174]

In a combustion of a hydrocarbon compound, 2 reactions are happening per element:

C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O

Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.

1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C

Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H

The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%

2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:

Amount of C = 0.01138 mol
Amount of H = 0.014244 mol

Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25

The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5

Thus, the empirical formula of the hydrocarbon is C₄H₅.

3. The molar mass of the empirical formula is

Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.

Molecular Formula = C₈H₁₀

4 0

3 years ago

Help me answer this question pls

Nostrana [21]

The answer should be D all of the above

3 0

3 years ago

Read 2 more answers

How do solve for #13?What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

exis [7]

What is the boiling point of a solution made by dissolving 1.0000 mole of sucrose in 1.0000 kg of water?

The change in Boiling Point of water can be calculated using this formula:

ΔTb = i * Kb * m

Where i is the van't hoff factor (the number of particles or ions), the kb is a constant (boiling point elevation constant) and m is the molality of the solution.

The kb for water is always 0.515 °C/m. Kb = 0.515 °C/m

The value for i in this case is 1. Since sucrose is a covalent compound and it doesn't dissociate into ions. i = 1

The molal concentration of the solution can be found using this formula:

molality = moles of sucrose/kg of water

molality = 1.000 mol / 1.000 kg of water

molality = 1 m

Now that we know all the values, we can use the formula to find the change in the boiling point of water:

ΔTb = i * Kb * m

ΔTb = 1 * 0.515 °C/m * 1 m

ΔTb = 0.515 °C

Finally, we are asked for the boiling point of the solution, not the change. The boiling point of water at atmospheric pressure is 100.00 °C. If the boiling point rises 0.515 °C when we prepare the solution. The boiling point of the solution is:

Boiling point solution = Boiling point of water + ΔTb

Boiling point solution = 100.000 °C + 0.515 °C

Boiling point solution = 100.515 °C

Answer: The boiling point of the solution is 100.515 °C.

8 0

10 months ago

How many moles of sulfate ions are in a sample of aluminum sulfate in which the number of formula units is 7.534 × 1023?

valentina_108 [34]

Answer:a unit of grammatical organization next below the sentence in rank and in traditional grammar said to consist of a subject and predicate

Explanation:

7 0

2 years ago