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Hunter-Best [27]
3 years ago
6

Some amount of hydrogen peroxide (H2O2) breaks down to produce 3 molecules of oxygen (O2) and 6 molecules of water (H2O). How ma

ny atoms of hydrogen are there?
Chemistry
2 answers:
cricket20 [7]3 years ago
8 0

Answer:

The answer to the question is ;

By counting the number of hydrogen in the decomposed hydrogen peroxide, there are a total 12 atoms of hydrogen in the reaction.

Explanation:

To solve the question, we note that

6H₂O₂ (l) → 6H₂O (l) + 3O₂ (g)

From the stoichiometry of the reaction, as it can be seen by balancing the equation that 6 moles of hydrogen peroxide liquid decomposes to produce 6 moles of liquid water molecule (H₂O) and 3 moles of oxygen gas molecules O₂.

Therefore, taking  count of hydrogen atoms in the reactant, there are 12 atoms of hydrogen atoms in the 6 molecules of hydrogen peroxide before the decomposition took place.

soldier1979 [14.2K]3 years ago
3 0

Answer:

Atoms_H=12Atoms

Explanation:

Hello,

In this case, the only source of hydrogen is in the 6 molecules of water, therefore, the atoms of hydrogen, by applying stoichiometry with the Avogadro's number is:

Atoms_H=6moleculesH_2O*\frac{1molH_2O}{6.022x10^{23}moleculesH_2O}\frac{2molH}{1molH_2O}*\frac{6.022x10^{23}AtomsH}{1molH} \\Atoms_H=12Atoms

Best regards.

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Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react
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Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of BCl_3 = 60 g

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Molar mass of BCl_3 = 117 g/mole

Molar mass of H_2O = 18 g/mole

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First we have to calculate the moles of BCl_3 and H_2O.

\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles

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Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)

From the balanced reaction we conclude that

As, 1 mole of BCl_3 react with 3 mole of H_2O

So, 0.513 moles of BCl_3 react with 3\times 0.513=1.539 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and BCl_3 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of HCl.

As, 1 mole of BCl_3 react to give 3 moles of HCl

So, 0.513 moles of BCl_3 react to give 3\times 0.513=1.539 moles of HCl

Now we have to calculate the mass of HCl.

\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl

\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g

Therefore, the theoretical yield of HCl is, 56.1735 grams

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