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Sauron [17]
3 years ago
5

What mass is 2.4E25 molecules of NaHCO3?

Chemistry
1 answer:
Zigmanuir [339]3 years ago
7 0

Answer:

Explanation:

Mole = no. Molecules/6.02×10^23

Mole = (2.4×10^25)/(6.02×10^23)

Mole = 39.87mole

Molar mass of NaHCO3 is

= 23 + 1 + 12 + 3(16)

= 84g/mol

Mole = mass/molar mass

Mass = Mole × molar mass

Mass = 39.87 × 84

Mass= 3348.84g

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5.943x10^24 molecules of H3PO4 will need how many grams of Mg(OH)2 in the reaction below? 3 Mg(OH)2 + 2 H3PO4 -------&gt; 1 Mg3(
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Answer:

Mass of Mg(OH)₂ required for the reaction = 863.13 g

Explanation:

3Mg(OH)₂ + 2H₃PO₄ -------> Mg₃(PO₄)₂ + 6H₂O

(5.943 x 10²⁴) molecules of H₃PO₄ is available fore reaction. Mass of Mg(OH)₂ required for reaction.

According to Avogadro's theory, 1 mole of all substances contain (6.022 × 10²³) molecules.

This can allow us find the number of moles that (5.943 x 10²⁴) molecules of H₃PO₄ represents.

1 mole = (6.022 × 10²³) molecules.

x mole = (5.943 x 10²⁴) molecules

x = (5.943 x 10²⁴) ÷ (6.022 × 10²³)

x = 9.87 moles

From the stoichiometric balance of the reaction,

2 moles of H₃PO₄ reacts with 3 moles of Mg(OH)₂

9.87 moles of H₃PO₄ will react with y moles of Mg(OH)₂

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So, 14.8 moles of Mg(OH)₂ is required for this reaction. We them convert this to mass

Mass = (number of moles) × Molar mass

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Hope this Helps!!!

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Answer: -

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