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Klio2033 [76]
3 years ago
5

Identify the particle that must receive 2 electrons to acquire a charge of +1. a) K b) Fe2+ c) O2- d) Nee) Al3+ (URGENT) Needs t

o be done in 30 mins
Chemistry
1 answer:
mrs_skeptik [129]3 years ago
5 0

Answer:

E) Al³⁺

Explanation:

A reaction involving a gain of electrons is known as a reduction reaction because the oxidation number of the species gaining the electron is reduced.

In the given question, the oxidation number (charge) of particle accepting two electrons will decrease by 2. From the given options;

A. K is a neutral atom with oxidation number of 0. If is accepts two electrons, its oxidation number becomes -2.

K + 2e⁻  ----> K⁻²

B) Fe²⁺ has a charge of +2. If it accepts two electrons, its charge comes 0.

Fe⁺ + 2e⁻  ----> Fe

C) O²⁻ has a charge of -2. if it accepts two electrons, it will have a charge of -4.

O²⁻ + 2e⁻  ---->  O⁴⁻

D) Ne has a charge of zero. If it accepts two electrons, its charge becomes -2.

Ne + 2e⁻   ---->   Ne²⁻

E) Al³⁺ has a charge of +3. If it gains two electrons, its charge becomes +1.

Al³⁺ + 2e⁻   ---->   Al⁺

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If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
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Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

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