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Tresset [83]
3 years ago
14

How many moles are there in 3.80x10^22 molecules of CCI4?

Chemistry
1 answer:
Elena-2011 [213]3 years ago
3 0

Answer:

0.0646 mole.

Explanation:

The following data were obtained from the question:

Number of molecules = 3.89×10²² molecules

Number of mole =?

The number of mole present in 3.89×10²² molecules of CCl₄ can be obtained as follow:

From Avogadro's hypothesis:

6.02×10²³ molecules = 1 mole

Therefore,

3.89×10²² molecules = 3.89×10²² molecules × 1 mole / 6.02×10²³ molecules

3.89×10²² molecules = 0.0646 mole

Thus, 0.0646 mole is present in 3.89×10²² molecules of CCl₄.

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The two half-reactions are...
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Let's start by balancing the first half-reaction...
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4H++NO3→NO+2H2O
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4e−+4H++NO3→NO+2H2O
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Sodium hydroxide reacts with aluminum and water to produce hydrogen gas: 2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H
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Answer:

The mass of hydrogen gas formed is 0.205 grams

Explanation:

<u>Step 1:</u> Data given

Mass of 1.83 grams of Al

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Molar mass of Al = 26.98 g/mol

Molar mass of NaOH = 40 g/mol

<u>Step 2:</u> The balanced equation:

2 Al(s) + 2 NaOH(aq) + 6 H2O(l) → 2 NaAl(OH)4(aq) + 3 H2(g)

<u>Step 3:</u> Calculate moles of Al

Moles Al = mass Al / Molar mass Al

Moles Al = 1.83 grams / 26.98 g/mol

Moles Al = 0.0678 moles

<u>Step 4:</u> Calculate moles of NaOH

Moles NaOH = 4.30 grams / 40 g/mol

Moles NaOH = 0.1075 moles

<u>Step 5</u>: Calculate limiting reactant

For 2 moles of Al, we need 2 moles of NaOH

Aluminium is the limiting reactant. It will completely be consumed ( 0.0678 moles)

NaOH is in excess. There will react 0.0678 moles

There will remain 0.1075 - 0.0678 = 0.0397 moles

<u>Step 6</u>: Calculate moles of hydrogen

For 2 moles of Al, we need 2 moles of NaOH, to produce 3 moles of hydrogen

For 0.0678 moles of Al, there is produced 0.0678 *3/2 = 0.1017 moles of H2

<u>Step 7</u>: Calculate mass of H2

Mass of H2 = Moles H2 * Molar mass of H2

Mass of H2 = 0.1017 moles * 2.02 g/mol

Mass of H2 = 0.205 grams

The mass of hydrogen gas formed is 0.205 grams

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