Answer:
In the given chemical reaction:
Species Oxidized: I⁻
Species Reduced: Fe³⁺
Oxidizing agent: Fe³⁺
Reducing agent: I⁻
As the reaction proceeds, electrons are transferred from I⁻ to Fe³⁺
Explanation:
Redox reaction is a chemical reaction involving the simultaneous movement of electrons thereby causing oxidation of one species and reduction of the other species.
The chemical species that <u><em>gets reduced by gaining electrons </em></u><u>is called an </u><u><em>oxidizing agent</em></u>. Whereas, the chemical species that <u><em>gets oxidized by losing electrons </em></u><u>is called a </u><u><em>reducing agent</em></u><u>.</u>
Given redox reaction: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
<u>Oxidation half-reaction</u>: 2 I⁻ + → I₂ + 2 e⁻ ....(1)
<u>Reduction half-reaction</u>: [ Fe³⁺ + 1 e⁻ → Fe²⁺ ] × 2
⇒ 2 Fe³⁺ + 2 e⁻ → 2 Fe²⁺ ....(2)
In the given redox reaction, <u>Fe³⁺ (oxidation state +3) accepts electrons and gets reduced to Fe²⁺ (oxidation state +2) and I⁻ (oxidation state -1) loses electrons and gets oxidized to I₂ (oxidation state 0).</u>
<u>Therefore, Fe³⁺ is the oxidizing agent and I⁻ is the reducing agent and the electrons are transferred from I⁻ to Fe³⁺.</u>
Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).
The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.
I hope this helps. Let me know in the comments if anything is unclear.
Answer:
Explanation:
C₂H₂ + 2H₂ = C₂H₆
1 mole 2 mole 1 mole
Feed of reactant is 1.6 mole H₂ / mole C₂H₂
or 1.6 mole of H₂ for 1 mole of C₂H₂
required ratio as per chemical reaction written above
2 mole of H₂ for 1 mole of C₂H₂
So H₂ is in short supply . Hence it is limiting reagent .
1.6 mole of H₂ will react with half of 1.6 mole or .8 mole of C₂H₂ to form .8 mole of C₂H₆
a )Calculate the stoichiometric reactant ratio = mole H₂ reacted/mole C₂H₂ reacted
= 1.6 / .8 = 2 .
b )
yield ratio = mole C₂H₆ formed / mole H₂ reacted ) = 0.8 / 1.6 = 1/2 = 0.5 .
Copper wire is not an example of a pure element because although it's made by pure elements, it's not one itself. It's made by factories.
Have a nice day! :)
