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KonstantinChe
2 months ago
6

An average copper penny minted in the 1960s contained about 3.000 g of copper. how much chalcopyrite had be mined to produce 100

pennies?
Chemistry
2 answers:
lorasvet [2.7K]2 months ago
8 0

To find the answer, start by calculating the total mass of the copper utilized:

Copper used = 100 pennies x 3.0g Cu per penny = 300.0 g Cu



Next, identify the path and molar ratios from Cu produced back to CuFeS2 needed using the established balanced reactions:

1 Cu2S from 2 CuS; 2Cu from 1 Cu2S; 2CuS from 2CuFeS2
Thus, 2Cu comes from 2CuFeS2, indicating a 1:1 molar ratio.



Then convert grams of Cu to moles and grams of CuFeS2:
= 300.0 g Cu * 1 mol Cu/63.546g Cu * 2 mol CuFeS2/2 moles Cu

= 4.72 moles CuFeS2



The required amount of chalcopyrite mined = 4.72 moles CuFeS2 * 183.54 g CuFeS2/1 mole CuFeS2 = 866.49 g CuFeS2

Anarel [2.9K]2 months ago
7 0

The total quantity of chalcopyrite that needs to be extracted to make 100 pennies is \boxed{{\text{866}}{\text{.49g}}}.

Additional clarification:

Stoichiometry examines the quantities of elements in reactions based on relationships among the involved reactants and products. This allows for the calculation of moles of a substance when the moles of other substances in the reaction are known.

Consider the basic reaction,

{\text{A}}+2{\text{B}}\to3{\text{C}}

In this equation,

A and B represent the reactants.

C indicates the product.

In this reaction, one mole of A reacts with two moles of B to form three moles of C. The stoichiometric ratios are 1:2 for A to B, 1:3 for A to C, and 2:3 for B to C.

Copper can be mined from naturally occurring forms of oxides or sulfides. To extract copper from chalcopyrite, a sequence of reactions is necessary. Initially, chalcopyrite is converted to copper sulfide following this reaction:

2{\text{CuFe}}{{\text{S}}_2}+3{{\text{O}}_2}\to2{\text{CuS}}+2{\text{FeO}}+2{\text{S}}{{\text{O}}_2}

In the subsequent step, the generated ferrous oxide is treated with silica as follows:

2{\text{FeO}}+{\text{Si}}{{\text{O}}_2}\to2{\text{FeSi}}{{\text{O}}_3}

In the tertiary step, the copper sulfide decomposes as presented below:

2{\text{CuS}}\to{\text{C}}{{\text{u}}_2}{\text{S}}+{\text{S}}

Finally, the resulting {\text{C}}{{\text{u}}_2}{\text{S}} undergoes oxidation to produce elemental copper.

{\text{C}}{{\text{u}}_2}{\text{S}}+{\text{S}}\to{\text{2Cu}}+2{\text{S}}{{\text{O}}_2}

These reactions show that 2 moles of {\text{CuFe}}{{\text{S}}_2} yield 2 moles of elemental copper. Therefore, the molar ratio of Cu to {\mathbf{CuFe}}{{\mathbf{S}}_{\mathbf{2}}} is 1:1.

Since a penny consists of 3 g of Cu, the total mass of copper in 100 pennies can be calculated as follows:

\begin{gathered}{\text{Mass of Cu}}=\left({\frac{{3\;{\text{g}}}}{{1\,{\text{penny}}}}}\right)\left({100\;{\text{pennies}}}\right)\\=300\;{\text{g}}\\\end{gathered}

Using the copper mass and its molar mass, we can find the moles of copper with this formula.

{\text{Number of moles}}=\frac{{{\text{Given mass}}}}{{{\text{Molar mass}}}} …… (1)

The mass provided is 300{\text{g}}.

The molar mass is {\text{63}}{\text{.546g/mol}}.

Insert the values into equation (1).

\begin{gathered}{\text{Number of moles}}=\frac{{300.0{\text{g}}}}{{{\text{63}}{\text{.546g/mol}}}}\\=4.72099{\text{mol}}\\\end{gathered}

Given the molar ratio of Cu and {\text{CuFe}}{{\text{S}}_2} being 1:1, the number of moles for {\mathbf{CuFe}}{{\mathbf{S}}_{\mathbf{2}}} is calculated to be 4.72099 mol.

The calculated moles of {\text{CuFe}}{{\text{S}}_2} can now lead us to the corresponding mass in grams. The expression for mass calculation is as follows:

{\text{Mass of CuFe}}{{\text{S}}_2}=\left({{\text{moles of CuFe}}{{\text{S}}_2}}\right)\left({{\text{molar mass of CuFe}}{{\text{S}}_2}}\right) …… (2)

The moles of {\text{CuFe}}{{\text{S}}_2} discovered is 4.72099{\text{ mol}}.

The molar mass of {\text{CuFe}}{{\text{S}}_2} is 183.54{\text{g/mol}}.

Substituting the values in equation (2):

\begin{gathered}{\text{Mass of CuFe}}{{\text{S}}_2}=\left({4.72099{\text{ mol}}}\right)\left({183.54{\text{g/mol}}}\right)\\=866.49{\text{g}}\\\end{gathered}

Learn more:

1. A statement regarding subatomic particles:

2. The net ionic equation for the reaction of {\text{MgS}}{{\text{O}}_{\text{4}}} with {\text{Sr}}{\left({{\text{N}}{{\text{O}}_3}}\right)_2}:

Details of the answer:

Grade: College

Subject: Chemistry

Chapter: Mole concept and stoichiometry

Keywords: copper, penny, pennies, CuFeS2, stoichiometry, O2, moles, mass, S and Cu2S.

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