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2 years ago

How much water would form if 4.8 grams of hydrogen reacted with 38.4 grams of oxygen?

1 answer:

3 0

Hydrogen reacts with oxygen to form water based on the following equation:
2H2 + O2 ..............> 2H2O
This means that 4 grams of hydrogen are required to react with 32 grams of oxygen to release 36 grams of water.
We need to calculate how munch oxygen is needed to react with 4.8 grams of hydrogen. We can do this using cross multiplication as follows:
amount of oxygen = (4.8x32) / 4 = 38.4 grams

This means that we have no limiting reagent here and both reactants will completely react.
We know that 4 grams of hydrogen are required to produce 36 grams of water. Therefore:
amount of water produced from 4.8 grams of hydrogen = (4.8 x 36) /4 = 43.2 grams

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The answer to this question is D! The ball and stick model! Hope this helps :)

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2 years ago

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Ethyne (C2 H2 (g), mc032-1.jpgHf = 226.77 kJ/mol) undergoes complete combustion in the presence of oxygen to produce carbon diox

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2 C2H2 + 5 02 > 4 CO2 + 2 H2O

Products - Reactants ( all units are kJ/mo1):

(4 x -393.5) + (2 x -241.82) - (2 x 226.77) - (5 x 0) = -2511.2 kJ/mo1

-2511.2 kJ/mo1 is for 2 moles of C2H2.The question asked for 1 mole of C2H2, so: -2511.2 / 2 = -1255.6 kJ/mo1

answer: -1255.6 kJ/mo1

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2 years ago

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Groups of atoms that are added to carbon backbones and give them unique properties are known as

Irina-Kira [14]

Answer:

Groups of atoms that are added to carbon backbones and give them unique properties are known as <u>Functional Groups</u>.

Explanation:

In organic chemistry they are called as Functional Group because they are the active part of a molecule. These groups give a unique characteristic to molecule both chemically and physically. Also, each functional group represent a different class of compounds.

Examples:

S No. Functional Group Name

1 R--X Alkyl Halides

2 R--OH Alcohols

3 R--NH₂ Amines

4 R--O--R Ethers

5 R--CO--R Ketones

6 R--CO--H Aldehydes

7 R--CO--OH Carboxylic acids

8 R--CO--X Acid Halides

10 R--CO--NR₂ Acid Amides

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2 years ago

the hydrogen gas generated when calcium metal reacts with water is collected over water at 20 degrees C. The volume of the gas i

Gala2k [10]

Answer:

There is 0.0677 grams of H2 gas obtained

Explanation:

Step 1: Data given

The total pressure (988 mmHg) is the sum of the pressure of the collected hydrogen + the vapor pressure of water (17.54 mmHg).

ptotal = p(H2)+ p(H2O)

p(H2) = ptotal - pH2O = 988 mmHg - 17.54 mmHg = 970.46 mmHg

Step 2: Calculate moles of H2 gas

Use the ideal gas law to calculate the moles of H2 gas

PV = nRT

n = PV / RT

⇒ with p = pressure of H2 in atm = 970.46 mmHg * (1 atm /760 mmHg) = 1.277 atm

⇒ V = volume of H2 in L = 641 mL x (1 L / 1000 mL) = 0.641 L

⇒ n = the number of moles of H2 = TO BE DETERMINED

⇒ R = the gas constant = 0.08206 L*atm/K*mol

⇒ T = the temperature = 20.0 °C = 293.15 Kelvin

n = (1.277)(0.641) / (0.08206)(298.15) = 0.0335 moles H2

Step 3: Calculate mass of H2

Mass of H2 = moles H2 ¨molar H2

0.0335 moles H2 * 2.02 g/mol H2 = 0.0677g H2

There is 0.0677 grams of H2 gas obtained

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2 years ago

50g nitrous oxide combines with 50g oxygen form dinitrogen tetroxide according to the balanced equation below.

photoshop1234 [79]

Limiting reactant : O₂

Mass of N₂O₄ produced = 95.83 g

<h3>Further explanation</h3>

Given

50g nitrous oxide

50g oxygen

Reaction

2N20 + 302 - 2N204

Required

Limiting reactant

mass of N204 produced

Solution

mol N₂O :

$\tt =\dfrac{50}{44}=1.136$

mol O₂ :

$\tt =\dfrac{50}{32}=1.5625$

2N₂O+3O₂⇒ 2N₂O₄

ICE method

1.136 1.5625

1.0416 1.5625 1.0416

0.0944 0 1.0416

Limiting reactant : Oxygen-O₂

Mass N₂O₄(MW=92 g/mol) :

$\tt =mol\times MW=1.0416\times 92=95.83~g$

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2 years ago