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OleMash [197]
1 year ago
10

Sixty-three grams of copper reacted with 32 grams of sulfur. this chemical reaction produced the compound copper sulfide. what w

as the mass of this compound?
Chemistry
1 answer:
Nuetrik [128]1 year ago
6 0

The mass of, Cu_2S, compound formed is 77.9g

62 grams of copper reacted with 32grams of sulfur to form copper sulfide.

Cu  +   S_2 \rightarrow  CuS_2

stoichiometry of Cu to S is 2:1

We need to find the limiting reactant

Molar mass of copper = 63.5 g/mol

Molar mass of Sulfur = 32 g/mol

Number of moles of Copper =  \dfrac{mass} {molar mass } = 0.99mole

Number of moles of Sulfur = \dfrac{mass} {molar mass} = \dfrac{32g} {32g/mol} = 1 mole

Since copper have lesser number of moles, therefore the limiting reagent is copper so the amount of product formed depends on amount of Cu present

stoichiometry of Copper to Cu_2S is 2:1

0.99 mol of Copper forms = \dfrac{0.99} {2}  = 0.49 mol of Cu_2S

Mass of Cu_2S produced = Number of moles  \times Molar mass                

Mass of Cu_2S produced = 0.49 mol \times 159 g/mol = 77.9g

Learn more about limiting reagent here-

brainly.com/question/9913056

#SPJ4

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makvit [3.9K]

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We are given:

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The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

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To calculate the E^o_{cell} of the reaction, we use the equation:

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How many molecules of fructose would you have if you have 0.7 moles of<br> fructose?<br> *
iogann1982 [59]

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4.214 × 10^23 molecules.

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