Chemistry, 02.02.2020 23:03 velaserik1

If a particular ore contains 58.5% calcium phosphate, what minimum mass of the ore must be processed to obtain 1.00 kg of phosphorus?

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Answer from: kaytlynnmeyer

The minimum mass of the ore is $\boxed{{\text{17094 g}}}$ .

Further explanation:

Molar mass:

The molar mass of a compound is the summation of masses of all atoms or molecules present in the 1 mole of compound. In other words, Molar mass of a compound is the ratio of the mass of compound in grams to the number of moles present of that compound. The S.I unit of molar mass is gram per mol or g/mol.

The molar mass of ${\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is calculated as follows:

${\text{Molar mass of}}{\mathbf{ }}{\text{C}}{{\text{a}}_{\text{3}}}{\left( {{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}=\left[\begin{gathered}{\text{3}}\left( {{\text{Atomic mass of Ca}}}\right)+\hfill\\{\text{2}}\left({{\text{Atomic mass of P}}}\right)+\hfill\\{\text{8}}\left( {{\text{Atomic mass of O}}} \right)\hfill\\\end{gathered}\right]$ …… (1)

The atomic mass of Ca is 40 g.

The atomic mass of P is 30.974 g.

The atomic mass of O is 16 g.

Substitute these values in equation (1).

$\begin{aligned}{\text{Molar mass of}}{\mathbf{}}{\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}} \right)_{\text{2}}}&=\left[{{\text{3}}\left({{\text{40 g}}} \right)+{\text{2}}\left({{\text{30}}{\text{.974 g}}}\right)+{\text{8}}\left({{\text{16 g}}} \right)}\right]\\&=\left[{{\text{120 g}}+{\text{61}}{\text{.948 g}}+{\text{128 g}}} \right]\\&=309.948\;{\text{g/mol}}\\&\approx {\text{310}}\;{\text{g/mol }}\\\end{aligned}$

The percentage of P in ${\mathbf{C}}{{\mathbf{a}}_{\mathbf{3}}}{\left( {{\mathbf{P}}{{\mathbf{O}}_{\mathbf{4}}}}\right)_{\mathbf{2}}}$ is calculated as follows:

${\text{\% of P}}=\left({\frac{{{\text{Atomic mass of P}}}}{{{\text{Molar mass of C}}{{\text{a}}_{\text{3}}}{{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}}\right)\left( {100\% }\right)$ …… (2)

The atomic mass of P is 30.974 g.

The atomic mass of ${\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is 309.948 g/mol.

Substitute these values in equation (2).

$\begin{aligned}{\text{\% of P}}&=\left({\frac{{{\text{30}}{\text{.974}}}}{{{\text{309}}{\text{.948}}}}}\right)\left({100\% }\right)\\&= 9.99328\;\%\\&\approx 10\;\%\\\end{aligned}$

The mass of phosphorus is converted to g. The conversion factor for this is,

${\text{1 kg}} = {\text{1}}{{\text{0}}^{\text{3}}}{\text{ g}}$

So the mass of phosphorus is calculated as follows:

$\begin{aligned}{\text{Mass of phosphorus}}&=\left({{\text{1 kg}}}\right)\left({\frac{{{\text{1}}{{\text{0}}^{\text{3}}}{\text{ g}}}}{{{\text{1 kg}}}}}\right)\\&={\text{1000 g}}\\\end{aligned}$

The amount of ${\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ is calculated as follows:

${\text{Amount of C}}{{\text{a}}_{\text{3}}}{\left( {{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}=\left({{\text{Mass of P}}}\right)\left({\frac{{{\text{100 \% of C}}{{\text{a}}_{\text{3}}}{{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}{{{\text{\% of P}}}}}\right)$ …… (3)

Substitute 1000 g for the mass of P and 10 % for % of P in equation (2).

$\begin{aligned}{\text{Amount of C}}{{\text{a}}_{\text{3}}}{\left( {{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}&=\left({{\text{1000 g}}}\right)\left({\frac{{{\text{100 \% }}}}{{{\text{10 \% }}}}}\right)\\&={\text{10000 g}}\\\end{aligned}$

The mass of ore can be calculated as follows:

${\text{Mass of ore}}=\left({\frac{{{\text{Mass of C}}{{\text{a}}_{\text{3}}}{{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}{{{\text{\% of C}}{{\text{a}}_{\text{3}}}{{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)}_{\text{2}}}}}}\right)\left({{\text{100 \% }}}\right)$ …… (4)

Substitute 10000 g for the mass of ${\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ and 58.5 % for % of ${\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}$ in equation (4).

$\begin{aligned}{\text{Mass of ore}}&=\left({\frac{{{\text{10000 g}}}}{{{\text{58}}{\text{.5 \% }}}}}\right)\left({{\text{100 \% }}}\right)\\&={\text{17094}}{\text{.017 g}}\\&\approx{\text{17094 g}}\\\end{aligned}$

So the minimum mass of the ore is 17094 g.

Learn more:

1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride:

2. Calculate the moles of ions in the solution:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Mole concept

Keywords: molar mass, P, atomic mass, 10 %, 58.5 %, 17094 g, ore, Ca3(PO4)2, 1000 g, 10000 g, Ca, P, O, atomic mass of Ca, atomic mass of P, atomic mass of O, conversion factor.

Answer from: Quest

why do the isotopes of a particular element behave differently in nuclear reactions but the same in chemical reactions? chemical reactions involve changes to electrons in an atom; isotopes do not differ in electrons, and so they react the same way during chemical changes.

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

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If a particular ore contains 58.5% calcium phosphate, what minimum mass of the ore must be processed...

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