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3 years ago

9

A neutral atom possesses an atomic number of 15 and an atomic mass of 31. Three electrons are gained. What is the result of this

conversion?

2 answers:

Katyanochek1 [597]3 years ago

6 0

<em>Answer:</em> an anion will form with -3 charge.

<em>Explanation;</em>

The atomic number of the given atom = 15

Atomic number = number of protons

The number of protons of the given atom = 15

The atomic mass of the given atom = 31

Atomic mass = number of protons + number of neutrons

Hence,

number of neutrons = atomic mass - number of protons

= 31 - 15

= 16

Since, the atomic number is 15, the given atom is P which has the atomic number as 15.

In the neutral compound,

number of protons = number of electrons

Hence,

the number of electrons of the neutral atom = 15

But, the atom has gained 3 electrons.

Gaining of electrons will form anions.

Hence, the atom will be converted into anion which has -3 charge on it.

But the atomic mass, atomic number, number of protons and number of electrons will not be changed

Romashka [77]3 years ago

3 0

Neutral atom carries atomic number 15 and atomic mass 31. This is phosphorous (P) atom with 15 electron. We know no. of proton are equal to no. of electron in neutral atom. After gain of 3 electrons, total number of electrons in P atom will change to 18. This results neutral atom onversion into a negative charge ion with a charge of 3-.

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Approximately $81.84\%$ .

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${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

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$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

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Also, when number of moles are equal in a solution then the formula will be as follows.

$M_{1} \times V_{1} = M_{2} \times V_{2}$

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