Respuesta :
Calcium Phospate formula: Ca3(PO4)2
Atomic weight:
Ca = 40 ; P = 31 ; O = 16
Ca3 = 40 * 3 = 120
P = 31
O4 = 16 * 4 = 64
(PO4)2 = (31 + 64) * 2 = 95 * 2 = 190
Ca3(PO4)2 = 120 + 190 = 310 g/mol
31/310 = 10% P in the calcium phospate
1.00kg * 1000 g/kg = 1000 g of phosphorus.
1000 g / 10% = 10,000 g of calcium phosphate
10,000 g / 58.5% = 17,094 grams of ore.
The minimum mass of the ore should be 17,094 grams or 17.094 kg.
Atomic weight:
Ca = 40 ; P = 31 ; O = 16
Ca3 = 40 * 3 = 120
P = 31
O4 = 16 * 4 = 64
(PO4)2 = (31 + 64) * 2 = 95 * 2 = 190
Ca3(PO4)2 = 120 + 190 = 310 g/mol
31/310 = 10% P in the calcium phospate
1.00kg * 1000 g/kg = 1000 g of phosphorus.
1000 g / 10% = 10,000 g of calcium phosphate
10,000 g / 58.5% = 17,094 grams of ore.
The minimum mass of the ore should be 17,094 grams or 17.094 kg.
The minimum mass of the ore is [tex]\boxed{{\text{17094 g}}}[/tex] .
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 [tex]{\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is calculated as follows:
[tex]{\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][/tex] …… (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).
[tex]\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}[/tex]
The percentage of P in [tex]{\mathbf{C}}{{\mathbf{a}}_{\mathbf{3}}}{\left( {{\mathbf{P}}{{\mathbf{O}}_{\mathbf{4}}}}\right)_{\mathbf{2}}}[/tex] is calculated as follows:
[tex]{\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)[/tex] …… (2)
The atomic mass of P is 30.974 g.
The atomic mass of [tex]{\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is 309.948 g/mol.
Substitute these values in equation (2).
[tex]\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}[/tex]
The mass of phosphorus is converted to g. The conversion factor for this is,
[tex]{\text{1 kg}} = {\text{1}}{{\text{0}}^{\text{3}}}{\text{ g}}[/tex]
So the mass of phosphorus is calculated as follows:
[tex]\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}[/tex]
The amount of [tex]{\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] is calculated as follows:
[tex]{\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)[/tex] …… (3)
Substitute 1000 g for the mass of P and 10 % for % of P in equation (2).
[tex]\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}[/tex]
The mass of ore can be calculated as follows:
[tex]{\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)[/tex] …… (4)
Substitute 10000 g for the mass of [tex]{\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] and 58.5 % for % of [tex]{\text{C}}{{\text{a}}_{\text{3}}}{\left({{\text{P}}{{\text{O}}_{\text{4}}}}\right)_{\text{2}}}[/tex] in equation (4).
[tex]\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}[/tex]
So the minimum mass of the ore is 17094 g.
Learn more:
1. Calculate the moles of chlorine in 8 moles of carbon tetrachloride: https://brainly.com/question/3064603
2. Calculate the moles of ions in the solution: https://brainly.com/question/5950133
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.
