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wel
3 years ago
10

When 2 grams of Hydrogen (H2) react with 32 grams of Oxygen (O2) the reaction yields hydrogen peroxide (H2O2). How many grams of

H2O2 are produced in the reaction?____
Chemistry
2 answers:
Alisiya [41]3 years ago
8 0

<u>Given information:</u>

Mass of H2 = 2 g

Mass of O2 = 32 g

<u>To determine:</u>

Mass of H2O2 produced

<u>Explanation:</u>

The reaction between H2 and O2 can be given as:

H2 + O2 → H2O2

Based on the reaction stoichiometry:

1 mole of H2 reacts with 1 mole of O2 to form 1 mole of H2O2

# moles of H2  = mass of H2 / molar mass of H2 = 2 g/ 2 g.mol-1 = 1 mole

# moles of O2 = mass of O2/ molar mass of O2 = 32 g/ 32 g.mol-1 = 1 mole

Hence for the given reactant conditions, moles of H2O2 produced = 1

Mass of H2O2 = moles of H2O2 * molar mass H2O2 = 1 mole * 34 g.mole-1 = 34 g

<u>Ans</u>: 34 g of H2O2 is produced in this reaction

harina [27]3 years ago
4 0
The answer is 34 grams

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First step is to get the mass of water molecule in grams:
From the periodic table:
molar mass of hydrogen is 1 
molar mass of oxygen is 16

molar mass of a water molecule = 2(1) + 16 = 18 gm

Now, to convert the gm into amu, all you have to do is multiply the gm you got by Avogadro's number as follows:
mass of water molecule = 18 x 6.22 x 10^23 = 1.1196 x 10^25 amu which is approximately 1 x 10^25 amu
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The spontaneous reaction that occurs when the cell in the picture operates is as follows: 2Ag+ + Cd(s) à 2 Ag(s) + Cd2+ (A) Volt
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The question is incomplete, the remaining part of the question is

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Answer:

Voltage decreases but remains > zero.

Explanation:

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How many grams of oxygen are required to burn 13.5 g of acetylene
Vinil7 [7]

Answer:

41.54 grams of oxygen are required to burn 13.5 g of acetylene

Explanation:

The balanced reaction is:

2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • C₂H₂: 2 moles
  • O₂: 5 moles
  • CO₂: 4 moles
  • H₂O: 2 moles

Being the molar mass of the compounds:

  • C₂H₂: 26 g/mole
  • O₂: 32 g/mole
  • CO₂: 44 g/mole
  • H₂O: 18 g/mole

By reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • C₂H₂: 2 moles* 26 g/mole= 52 grams
  • O₂: 5 moles* 32 g/mole= 160 grams
  • CO₂: 4 moles* 44 g/mole= 176 grams
  • H₂O: 2 moles* 18 g/mole= 36 grams

You can apply the following rule of three: if by stoichiometry 52 grams of acetylene react with 160 grams of oxygen, 13.5 grams of acetylene react with how much mass of oxygen?

mass of oxygen=\frac{13.5 grams of acetylene*160 grams of oxygen}{52 grams of acetylene}

mass of oxygen= 41.54 grams

<u><em>41.54 grams of oxygen are required to burn 13.5 g of acetylene</em></u>

<u><em></em></u>

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tatyana61 [14]

Answer:

Hypotheses

Explanation:

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