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ahrayia [7]
3 years ago
11

How many moles of oxygen are required for the combustion of 35.5 g of Magnesium Mg+O2-->MgO Please balance

Chemistry
1 answer:
Verdich [7]3 years ago
6 0

Answer:

0.74 moles of oxygen are required for combustion

Explanation:

Given data:

Number of moles of oxygen required = ?

Mass of Mg = 35.5 g

Solution:

Chemical equation:

2Mg + O₂         →       2MgO

Number of moles of Mg:

Number of moles = mass/molar mass

Number of moles = 35.5 g/ 24 g/mol

Number of moles = 1.48 mol

Now we will compare the moles of Mg with oxygen.

                  Mg           :         O₂

                   2             :          1

                 1.48           :        1/2×1.48 = 0.74 mol

0.74 moles of oxygen are required for combustion.

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When a 17.7 mL sample of a 0.368 M aqueous hypochlorous acid solution is titrated with a 0.301 M aqueous barium hydroxide soluti
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Answer:

pH = 12.98

Explanation:

Step 1: Data given

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Molarity of aqueous hypochlorous acid solution = 0.368 M

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Volume of aqueous barium hydroxide solution = 16.2 mL = 0.0162 L

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Step 3: Calculate moles

Moles = molarity * volume

Moles HCl = 0368 M * 0.0177 L

Moles HCl = 0.0065136 moles

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Moles Ba(OH)2 = 0.0048762 moles

Step 4: Calculate the limiting reactant

For 2 moles HCl we need 1 mol Ba(OH)2 to produce 1 mol BaCl2 and 2 moles H2O

HCl is the limiting reactant. It will completely be consumed 0.0065136 moles. Ba(OH)2 is in excess. There will react 0.0065136/2 = 0.0032568‬ moles. There will remain 0.0048762 moles - 0.0032568‬  = 0.0016194 moles

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Molarity Ba(OH)2 = moles / volume

Molarity Ba(OH)2 = 0.0016194 moles / 0.0339 L

Molarity Ba(OH)2 = 0.04777 M

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For Ba(OH)2 we have 2* [OH-]

[OH-] = 2*0.04777 = 0.09554 M

Step 7: Calculate pOH

pOH = -log[OH-]

pOH = -log(0.09554)

pOH = 1.02

Step 8: Calculate pH

pH = 14 - 1.02

pH = 12.98

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