Which of the following will reduce copper? zinc mercury fluorine chlorine

Chemistry

2 answers:

garri49 [273]4 years ago

6 0

<u>Answer:</u> The correct answer is Zinc.

<u>Explanation:</u>

Reactivity is defined as the tendency of an element to loose or gain electrons. Metals which loose electrons easily will be considered as more reactive metal.

The reactivity of elements will be judged on the basis of reactivity series. The elements which lie above in the series are more reactive than the elements which lie below in the series. The metals which lie above in the series will easily reduce the metals which lie below in the series.

Copper is the metal which lie above mercury and below zinc.

Thus, zinc is more reactive metal than copper and hence will easily reduce copper from its reaction.

Hence, the correct answer is Zinc.

laiz [17]4 years ago

4 0

Zinc because the only metals that would be able to reduce copper ions in solution would be hydrogen, lead, tin, nickel, iron, zinc, aluminum, Magnesium, sodium, calcium, potassium, and lithium. and according to your answer choices Zinc is the answer.

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<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + 2 O₂ → CO₂ + 2 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

CH₄: 1 mole
O₂: 2 moles
CO₂: 1 mole
H₂O: 2 moles

The molar mass of the compounds is:

CH₄: 16 g/mole
O₂: 32 g/mole
CO₂: 44 g/mole
H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

CH₄: 1 mole ×16 g/mole= 16 grams
O₂: 2 moles ×32 g/mole= 64 grams
CO₂: 1 mole ×44 g/mole= 44 grams
H₂O: 2 moles ×18 g/mole=36 grams

<h3>Limiting reagent</h3>

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

<h3>Limiting reagent in this case</h3>

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 16 grams of CH₄ reacts with 64 grams of O₂, 8.50 grams of CH₄ reacts with how much mass of O₂?

$mass of O_{2} =\frac{8.50 grams of CH_{4}x64 grams of O_{2} }{16grams of CH_{4}}$

mass of O₂= 34 grams

But 34 grams of O₂ are not available, 15.9 grams are available. Since you have less mass than you need to react with 8.50 grams of CH₄, O₂ will be the limiting reagent.

<h3>Mass of CO₂ formed</h3>

The following rule of three can be applied: if by reaction stoichiometry 64 grams of O₂ form 44 grams of CO₂, 15.9 grams of O₂ form how much mass of CO₂?

$mass of CO_{2} =\frac{15.9 grams of O_{2}x44 grams of CO_{2} }{64grams of O_{2}}$