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Harlamova29_29 [7]
2 years ago
11

given that the molar mass of Na2SO4 × nH2O is 322.1 g/mol calculate the number of moles of water in this hydrate

Chemistry
1 answer:
Masja [62]2 years ago
7 0
The molar mass of anhydrous Na₂SO₄ = [2(23) + 32 + 4(16)]
                                                                = 142 g/mol

If the molar mass of hydrated Na₂SO₄ = 322 g/mol

Since hydrated salt = Anhydrous Salt + Water

Then the mass of hydration (Water)= 322 g/mol - 142 g/mol    
                                                           = 180 g/mol

Since the molar mass of water itself is 18 g/mol
⇒ Moles of water = 180 g/mol ÷ 18g/mol
                              = 10

Thus n = 10; the formula of the hydrated salt is Na₂SO₄ · 10H₂O (Sodium Sulphate Decahydrate)
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The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare a 0.500 M HCl(aq) solution i
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Answer:

110ml

Explanation:

<em>Using the dilution equation, C1V1 = C2V2</em>

<em>Where C1 is the initial concentration of solution</em>

<em>C2 is final concentration of solution</em>

<em>V1 is intital volume of solution</em>

<em>V2 is final volume of solution.</em>

From the question , C1=6M, C2=0.5M, V1=10ml, V2=?

V2 =\frac{C1V1}{C2}

V2 =\frac{10*6}{0.5}

V2 =120ml

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3 0
2 years ago
A 8.6 g sample of methane and 15.6 g sample of oxygen react according to the reaction in the video. identify the limiting reacta
GalinKa [24]
Answer:

<span>23.6 g carbon dioxide comes from 8.6 g of CH4 or 10.7 g carbon dioxide comes from 15.6 g O that means the 15.6 g of oxygen is still the limiting reactant because it gets used up and only makes 10.7 g of CO2. </span>

Explanation:

1) Balanced chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

2) mole ratios:
1 mol CH₄ : 2mol O₂ : 1 mol CO₂ : 2 mol H₂O

3) molar masses
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O₂: 32.0 g/mol
CO₂: 44.01 g/mol

4) Convert the reactant masses to number of moles, using the formula 

number of moles = mass in grams / molar mass


CH₄: 8.6g / 16.04 g/mol = 0.5362 moles
<span />

O₂: 15.6 g / 32.0 g/mol = 0.4875 moles

5) If the whole 0.5632 moles of CH₄ reacted that yields to the same number of moles of CO₂ and that is a mass of:
mass of CO₂ = number of moles x molar mass = 23.60 g of CO₂

Which is what the first part of the answer says.

6) If the whole 0.4875 moles of O₂ reacted that would yield 0.4875 / 2 = 0.24375 moles of CO₂, and that is a mass of:
mass of CO₂ = 0.4875 grams x 44.01 g/mol = 10.7 grams of CO₂.

Which is what the second part of the answer says.

7) From the mole ratio you know infere that 0.5362 moles of CH₄ needs more twice number of moles of O₂, that is 1.0724 moles of O₂, and since there are only 0.4875 moles of O₂, this is the limiting reactant.

Which is what the chosen answer says.

8) From the mole ratios 0.4875 moles of O₂ produce 0.4875 / 2 moles of CO₂, and that is:
0.4875 / 2 mols x 44.01 g/mol = 10.7 g of CO₂, which is the last part of the answer.

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Answer:

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