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Marat540 [252]
3 years ago
15

The balanced equation for water is 2 H2 + O2 to 2 H2O. If I have 21.2g of a product , and I started with 5.6 g of H2, how many g

rams of oxygen were reacted
Chemistry
1 answer:
ANTONII [103]3 years ago
7 0

Since 21.2 g H2O was produced, the amount of oxygen that reacted can be obtained using stoichiometry. The balanced equation was given:  2H₂ + O₂ → 2H₂O and the molar masses of the relevant species are also listed below. Thus, the following equation is used to determine the amount of oxygen consumed.

Molar mass of H2O = 18 g/mol

Molar mass of O2 = 32 g/mol

21.2 g H20 x 1 mol H2O/ 18 g H2O x 1 mol O2/ 2 mol H2O x 32 g O2/ 1 mol O2 = 18.8444 g O2

<span>We then determine that 18.84 g of O2 reacted to form 21.2 g H2O based on stoichiometry. It is important to note that we do not need to consider the amount of H2 since we can derive the amount of O2 from the product. Additionally, the amount of H2 is in excess in the reaction.</span>

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Be sure to answer all parts. Write the equations representing the following processes. In each case, be sure to indicate the phy
barxatty [35]

Answer:

1.S^{-}(g)+e^{-} \rightarrow S^{2-}(g)

2.Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}

3.The electron affinity of  Mg^{2+} is zero.

4.O^{2-}(g) \rightarrow O^{-}(g)+e^{-}

Explanation:

1.

<u>Electron affinity:</u>

It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.

The electron affinity of S^{-} is as follows.

S^{-}(g)+e^{-} \rightarrow S^{2-}(g)

2.

<u>Ionization energy</u>:

Amount of energy required to removal of an electron from an isolated gaseous atom.

The third ionization energy of Titanium is as follows.

Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}

3.

The electronic configuration of Mg: 1s^{2}2s^{2}2p^{6}3s^{2}

By the removal of two electrons from a magnesium element we get Mg^{2+} ion.

Mg^{2+} has inert gas configuration i.e,1s^{2}2s^{2}2p^{6}

Hence, it does not require more electrons to get stability.

Therefore,the electron affinity of  Mg^{2+} is zero.

4.

The ionization energy of O^{2-} is follows.

O^{2-}(g) \rightarrow O^{-}(g)+e^{-}

3 0
3 years ago
For each of the following substituents, indicate whether it withdraws electrons inductively, donates electrons by hyperconjugati
aleksandrvk [35]

Answer:

Br- Withdraws electrons inductively

       Donates electrons by resonance

CH2CH3 - Donates electrons by hyperconjugation

NHCH3- Withdraws electrons inductively

              Donates electrons by resonance

OCH3 -  Withdraws electrons inductively

              Donates electrons by resonance

+N(CH3)3 - Withdraws electrons inductively

                   

Explanation:

A chemical moiety may withdraw or donate electrons by resonance or inductive effect.

Halogens are electronegative elements hence they withdraw electrons by inductive effect. However, they also contain lone pairs so the can donate electrons by resonance.

Alkyl groups donate electrons by hyperconjugation involving hydrogen atoms.

-NHCH3  and contain species that have lone pair of electrons which can be donated by resonance. Also, the nitrogen and oxygen atoms are very electron withdrawing making the carbon atom to have a -I inductive effect.

+N(CH3)3 have no lone pair and is strongly electron withdrawing by inductive effects.

3 0
3 years ago
If 28 grams of N reacts completely with 12 grams of H2, then how many
Bogdan [553]

Answer:

Mass of NH₃ produced = 34 g

Explanation:

Given data:

Mass of nitrogen = 28 g

Mass of Hydrogen = 12 g

Mass of NH₃ produced = ?

Solution:

Chemical equation:

N₂ +  3H₂    →   2NH₃

Moles of nitrogen:

Number of moles = mass/molar mass

Number of moles = 28 g/ 28 g/mol

Number of moles = 1 mol

Moles of hydrogen:

Number of moles = mass/molar mass

Number of moles = 12 g/ 2 g/mol

Number of moles = 6 mol

Now we will compare the moles of hydrogen and nitrogen with ammonia.

                            H₂              :               NH₃

                            3                :                2

                            6                :             2/3×6 = 4 mol

                           N₂              :                NH₃

                            1                :                 2

Number of moles of ammonia produced by nitrogen are less thus it will act as limiting reactant.

Mass of ammonia produced:

Mass = number of moles × molar mass

Mass =  2 mol  ×  17 g/mol

Mass = 34 g

                     

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