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Vinvika [58]
2 years ago
14

A sample of 85.5 g of tetraphosphorous decoxide (P4O10) reacts with 74.9 g of water to produce phosphoric acid (H3PO4) according

to the following balanced equation.
P4O10+6H2O⟶4H3PO4

Determine the limiting reactant for the reaction.

H2O

H3PO4

P4O10
Calculate the mass of H3PO4 produced in the reaction.
mass of H3PO4:
g
Calculate the percent yield of H3PO4 if 39.2 g of H3PO4 is isolated after carrying out the reaction.
percent yield:
%
Chemistry
1 answer:
Black_prince [1.1K]2 years ago
8 0

The limiting reactant here is P4O10 . The percent yield of the product is caculated as 33.3%.

<h3>What is the limiting reactant?</h3>

The limiting  reactant is the reactant that is in a minute quantity.

Number of moles of P4O10 =  85.5 g/284 g/mol = 0.3 moles

Number of moles of H2O =  74.9 g / 18 g/mol = 4.2 moles

From the reaction equation;

1 mole of P4O10 reacts with 6 moles of H2O

x moles of P4O10 reacts with 4.2 moles of H2O

x = 0.7  moles

Hence, P4O10 is the limiting reactant.

1 mole of P4O10 yields 4 moles of H3PO4

0.3 moles of P4O10 yields

0.3 moles * 4/1 = 1.2 moles

Mass of the H3PO4 = 1.2 moles * 98 g/mol = 117.6 g

Percent yield = 39.2 g/117.6 g * 100/1 = 33.3 %

Learn more about limiting reactant:brainly.com/question/14225536

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When a gas is cooled the particles is moving fast or slowly
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Answer:slowly

Explanation: when a gas is being cooled then its particles are gonna slowly stop moving to form a liquid

4 0
2 years ago
CHEMISTRY HELP PLEASE *answer all questions please*
Lerok [7]

Answer:

<u><em>Question 1: </em></u>

A) 0.289 moles.

B) 1.74 x 10²³ atoms.

<u><em>Question 2:</em></u>

A) 0.30 moles.  

B) it contains 0.3 moles of both Na and Cl.

C) it contains 6.023 x 10²³ atoms of both Na and Cl.

<u><em>Question 3:</em></u>

A) The number of moles of sucrose (C₁₂H₂₂O₁₁) ≅ 0.0228 moles.

B) The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2763 mole of C atoms.

The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = 0.5016 mole of H atoms.

The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = 0.2508 mole of O atoms.

C) The number of C atoms = 1.65 x 10²³ atoms.

The number of H atoms = 3.02 x 10²³ atoms.

The number of O atoms = 1.51 x 10²³ atoms.

Explanation:

<u><em>Question 1:</em></u>

A) The number of moles of Au in 57.01 g sample:

n = mass / molar mass,

mass = 57.01 g and molar mass = 196.966 g/mol e.

The number of moles of Au in the sample = (57.01 g) / (196.966 g/mole) = 0.289 moles.

B) The number of atoms of Au in the sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of Au → 6.023 x 10²³ atoms

0.289 mole of Au → ???? atoms

<em>using cross multiplication: </em>

The number of atoms of Au in the sample = (6.023 x 10²³ x 0.289 mole) / (1.0 mole) = 1.74 x 10²³ atoms.


<u><em>Question 2:</em></u>

A) The number of moles of 17.45 g of NaCl:

n = mass / molar mass,

mass = 17.45 g and molar mass = 58.44 g/mole.

The number of moles of NaCl = (17.45 g) / (58.44 g/mole) = 0.298 mole ≅ 0.30 moles.

B) The number of moles of each element in NaCl  

NaCl → Na + Cl

Each mole of NaCl contains one mole of Na and one mole of Cl.

<em><u>using cross multiplication: </u></em>

1.0 mole NaCl → 1.0 mole Na

0.3 mole NaCl → ??? mole Na

The number of moles of Na atoms in NaCl = (1.0 mole Na x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Na atoms.

by the same way; the number of moles of Cl atoms = (1.0 mole Cl x 0.3 mole NaCl) / (1.0 mole NaCl) = 0.3 mole of Cl atoms.

C) The number of atoms of each element in the sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of NaCl → 6.023 x 10²³ molecules

0.3 mole of NaCl → ???? molecules

<em><u>using cross multiplication:</u></em>

The number of molecules in 0.3 mole of NaCl = (6.023 x 10²³ x 0.3 mole) / (1.0 mole) = 1.8069 x 10²³ molecules.

Every molecule of NaCl contains one atom of Na and one atom of Cl.

So, it contains 6.023 x 10²³ atoms of both Na and Cl.


<u><em>Question 3:</em></u>

A) The number of moles of 7.801 g of sucrose (C₁₂H₂₂O₁₁):

n = mass / molar mass,

mass = 7.801 g and molar mass = 342.3 g/mole.

The number of moles of sucrose (C₁₂H₂₂O₁₁) = (7.801 g) / (342.3 g/mol) = 0.022789 mol ≅ 0.0228 moles.

B) The number of moles of each element in sucrose (C₁₂H₂₂O₁₁):

C₁₂H₂₂O₁₁ → 12C + 22H + 11O

Each mole of sucrose contains 12 moles of C, 22 moles of H, and 11 moles of O.

  • <em><u>using cross multiplication: </u></em>

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 12.0 moles C

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles C

The number of moles of C atoms in sucrose (C₁₂H₂₂O₁₁) = (12.0 moles C x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2763 mole of C atoms.

  • By the same way; the number of moles of H atoms:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 22.0 moles H

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles H

The number of moles of H atoms in sucrose (C₁₂H₂₂O₁₁) = (22.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.5016 mole of H atoms.

  • Also; the number of moles of O atoms:

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 11.0 moles O

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ??? moles O

The number of moles of O atoms in sucrose (C₁₂H₂₂O₁₁) = (11.0 moles H x 0.0228 moles of sucrose (C₁₂H₂₂O₁₁)) / (1.0 mole sucrose (C₁₂H₂₂O₁₁)) = 0.2508 mole of O atoms.

C) The number of atoms of each element in the sucrose (C₁₂H₂₂O₁₁) sample:

It is known that every mole of a substance contains Avogadro,s number (NA = 6.023 x 10²³) of molecules.

1.0 mole of sucrose (C₁₂H₂₂O₁₁) → 6.023 x 10²³ molecules

0.0228 mole of sucrose (C₁₂H₂₂O₁₁) → ???? molecules

<em><u>using cross multiplication: </u></em>

The number of molecules in 0.0228 mole of sucrose (C₁₂H₂₂O₁₁) = (6.023 x 10²³ x 0.0228 mole) / (1.0 mole) = 1.273 x 10²² molecules.

Each molecule of sucrose contains 12 atoms of C, 22 atoms of H, and 11 atoms of O.

So, the number of each atom that the sucrose (C₁₂H₂₂O₁₁) sample contains are:

The number of C atoms = (12 x 1.273 x 10²² molecules) = 1.65 x 10²³ atoms.

The number of H atoms = (22 x 1.273 x 10²² molecules) = 3.02 x 10²³ atoms.

The number of O atoms = (11 x 1.273 x 10²² molecules) = 1.51 x 10²³ atoms.

6 0
2 years ago
In the formula for carbon tetrahydride, CH4, the percent composition of all elements is:
creativ13 [48]
C%=12/16*100=3/4*100=75%
H%=25%
8 0
2 years ago
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8.43g of CaCl2·2H2O is dissolved in 40.0g of H2O. What is the molality of the solution? (Note that the H2O in the CaCl2·2H2O mus
lara31 [8.8K]

Answer:

[CaCl₂·2H₂O] = 1.43 m

Explanation:

Molality is mol of solute / kg of solvent.

Mass of solvent = 40 g

Let's convert g to kg → 40 g / 1000 = 0.04 kg

Let's determine the moles of solute (mass / molar mass)

8.43 g / 146.98 g/mol = 0.057 mol

Molality = 0.057 mol / 0.04 kg → 1.43

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