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Rama09 [41]
2 years ago
8

You use 4.98 g of Na2SO4, how many grams of barium sulfate are produced? Na2+Ba(NO3)2= ?

Chemistry
1 answer:
Kaylis [27]2 years ago
6 0

Answer:

Mass of barium sulfate = 8.17 g

Explanation:

Given data:

Mass of sodium sulfate = 4.98 g

Mass of barium sulfate produced = ?

Solution:

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

Moles of sodium sulfate:

Number of moles = mass/molar mass

Number of moles =4.98 g / 142.04 g/mol

Number of moles = 0.035 mol

Now we will compare the moles pf sodium sulfate and with barium sulfate.

                            Na₂SO₄               :            BaSO₄

                                1                       :              1

                              0.035                :          0.035

Mass of barium sulfate:

Mass = number of moles × molar mass

Mass =   0.035 mol ×233.4 g/mol

Mass = 8.17 g

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The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:

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1 - 0.115 = 0.885 mol CH₂Cl₂

We need moles of solute, which we have, and must convert our moles of solvent to kg:

0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂

We can now calculate the molality:

m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂

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