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Chapter 3: Problem 133

Bornite \(\left(\mathrm{Cu}_{3} \mathrm{Fe} \mathrm{S}_{3}\right)\) is a copper ore used in the production of copper. When heated, the following reaction occurs: $$ 2 \mathrm{Cu}_{3} \mathrm{FeS}_{3}(s)+7 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{Cu}(s)+2 \mathrm{FeO}(s)+6 \mathrm{SO}_{2}(g) $$ If 2.50 metric tons of bornite is reacted with excess \(\mathrm{O}_{2}\) and the process has an 86.3\(\%\) yield of copper, what mass of copper is produced?

Short Answer

Expert verified
The mass of copper produced is: $$ \frac{2.50 × 10^6\,\text{grams}}{(3×63.5) + (55.8) + (3×32.1)\,\text{grams/mol}} × \frac{6\,\text{moles of Cu}}{2\,\text{moles of bornite}} × 63.5\,\text{grams/mol} × \frac{86.3\%}{100} = 1.81 × 10^6\,\text{grams} $$ The actual mass of copper produced is approximately 1.81 metric tons.

Step by step solution

01

Convert the mass of bornite into moles

To find the number of moles of bornite, we need to divide the mass of bornite by its molar mass. The mass of bornite is 2.50 metric tons, which is equivalent to 2.50 × 10\(^6\) grams. The molar mass of \(\mathrm{Cu}_{3} \mathrm{FeS}_{3}\) can be calculated as follows: \([(3×63.5) + (1×55.8) + (3×32.1)]\) g/mol, where 63.5 g/mol is the molar mass of copper, 55.8 g/mol is the molar mass of iron, and 32.1 g/mol is the molar mass of sulfur.
02

Calculate the number of moles of bornite

Divide the mass of bornite by its molar mass to find the number of moles of bornite: $$ \text{moles of bornite} = \frac{2.50 × 10^6\,\text{grams}}{(3×63.5) + (55.8) + (3×32.1)\,\text{grams/mol}} $$
03

Use stoichiometry to calculate the moles of copper produced

According to the balanced chemical equation, 2 moles of bornite produces 6 moles of copper. To find the moles of copper produced from the reaction, we can use the ratio of moles of bornite to moles of copper: $$ \text{moles of Cu} = \text{moles of bornite} × \frac{6\,\text{moles of Cu}}{2\,\text{moles of bornite}} $$
04

Convert moles of copper to mass

To convert the moles of copper produced to mass, we will multiply the moles with the molar mass of copper: $$ \text{mass of Cu} = \text{moles of Cu} × 63.5\,\text{grams/mol} $$
05

Apply the percentage yield

The process has an 86.3% yield of copper. To calculate the actual mass of copper produced, multiply the theoretical mass of copper by the percentage yield and divide by 100: $$ \text{actual mass of Cu} = \text{mass of Cu} × \frac{86.3\%}{100} $$
06

Calculate the mass of copper produced

Calculate the actual mass of copper produced based on the above steps. This will give the final answer to the problem.

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Most popular questions from this chapter

A 9.780 -g gaseous mixture contains ethane $\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\( and propane \)\left(\mathrm{C}_{3} \mathrm{H}_{8}\right) .$ Complete combustion to form carbon dioxide and water requires 1.120 \(\mathrm{mole}\) of oxygen gas. Calculate the mass percent of ethane in the original mixture.

A compound contains only \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{N}\) . Combustion of 35.0 \(\mathrm{mg}\) of the compound produces 33.5 $\mathrm{mg} \mathrm{CO}_{2}\( and 41.1 \)\mathrm{mg}\( \)\mathrm{H}_{2} \mathrm{O}$ . What is the empirical formula of the compound?

Elixirs such as Alka-Seltzer use the reaction of sodium bicarbonate with citric acid in aqueous solution to produce a fizz: $$ 3 \mathrm{NaHCO}_{3}(a q)+\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}(a q) \longrightarrow $$ $$ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(a q) $$ a. What mass of \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7}\) should be used for every \(1.0 \times 10^{2} \mathrm{mg} \mathrm{NaHCO}_{3} ?\) b. What mass of \(\mathrm{CO}_{2}(g)\) could be produced from such a mixture?

Glass is a mixture of several compounds, but a major constituent of most glass is calcium silicate, \(\mathrm{CaSiO}_{3}\) . Glass can be etched by treatment with hydrofluoric acid; HF attacks the calcium silicate of the glass, producing gaseous and water-soluble products (which can be removed by washing the glass). For example, the volumetric glassware in chemistry laboratories is often graduated by using this process. Balance the following equation for the reaction of hydrofluoric acid with calcium silicate. $$ \mathrm{CaSiO}_{3}(s)+\mathrm{HF}(a q) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$

In using a mass spectrometer, a chemist sees a peak at a mass of 30.0106 . Of the choices $^{12} \mathrm{C}_{2}^{1} \mathrm{H}_{6},^{12} \mathrm{C}^{1} \mathrm{H}_{2}^{16} \mathrm{O},\( and \)^{14} \mathrm{N}^{16} \mathrm{O}$ which is responsible for this peak? Pertinent masses are \(^{1} \mathrm{H}\) $1.007825 ; 16 \mathrm{O}, 15.994915 ;\( and \)^{14} \mathrm{N}, 14.003074$
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