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Marina86 [1]
3 years ago
12

Can the excess reactant control the percentage yield

Chemistry
1 answer:
katrin2010 [14]3 years ago
6 0

Explanation:

when one reactants is in excess, there will always be some left over. The other reactants becomes limiting factor and controls show much of each product is produced.while using excess percentage yields this is at the expense of atom economy.

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hooc are carboxyl groups

your r or amino groups are those unique structures which have different atoms in them. your nh2 groups are your hydrogen atoms

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the scientific law of conservation of mass states that the total amount of mass in an isolated system remains constant. This mea
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It describes how, when particles/mc/elements react, despite forming different substances the mass is neither created nor destroyed but only converted.
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What are the correct half reactions for the following reaction:
d1i1m1o1n [39]

Answer:

D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.

Explanation:

  • It is a redox reaction that is consisted of two half-reactions:

Oxidation reaction:

Zn losses 2 electrons and is oxidized to Zn²⁺:

<em>Zn → Zn²⁺ + 2e⁻.</em>

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Reduction reaction:

H⁺ gains 1 electron and is reduced to H:

<em>2H⁺ + 2e⁻ → H₂.</em>

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<em>So, the right choice is: D. Zn → Zn²⁺ + 2e⁻, 2H⁺ + 2e⁻ → H₂.</em>

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What causes the wind? A.The Earths orbit around the sun B.The Earths rotation on its axis C.The layers of the Earths atmosphere
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Uneaven hearing by the sun

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What mass of Fe(OH)3 is produced when 35 mL of 0.250 M Fe(NO3)3 solution is mixed with 55 mL of a 0.180 M
Zina [86]

Answer:

0.35 g.

Explanation:

We'll begin by calculating the number of mole of Fe(NO3)3 in 35 mL of 0.250 M Fe(NO3)3 solution.

This is illustrated below:

Molarity of Fe(NO3)3 = 0.250 M

Volume = 35 mL = 35/1000 = 0.035 L

Mole of Fe(NO3)3 =?

Molarity = mole /Volume

0.250 = mole of Fe(NO3)3 / 0.035

Cross multiply

Mole of Fe(NO3)3 = 0.25 x 0.035

Mole of Fe(NO3)3 = 8.75×10¯³ mole

Next, we shall determine the number of mole of KOH in 55 mL of 0.180 M

KOH solution. This is illustrated below:

Molarity of KOH = 0.180 M

Volume = 55 mL = 55/1000 = 0.055 L

Mole of KOH =.?

Molarity = mole /Volume

0.180 = mole of KOH /0.055

Cross multiply

Mole of KOH = 0.180 x 0.055

Mole of KOH = 9.9×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

3KOH + Fe(NO3)3 —> Fe(OH)3 + 3KNO3

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3 to produce 1 mole of Fe(OH)3.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted with 1 mole of Fe(NO3)3.

Therefore, 9.9×10¯³ mole of KOH will react with = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(NO3)3.

From the above illustration, we can see that only 3.3×10¯³ mole out of 8.75×10¯³ mole of Fe(NO3)3 given is needed to react completely with 9.9×10¯³ mole of KOH.

Therefore, KOH is the limiting reactant and Fe(NO3)3 is the excess reactant.

Next, we shall determine the number of mole of Fe(OH)3 produced from the reaction.

In this case, we shall use the limiting reactant because it will give the maximum yield of Fe(OH)3 as all of it is consumed in the reaction.

The limiting reactant is KOH and the mole of Fe(OH)3 produce can be obtained as follow:

From the balanced equation above,

3 moles of KOH reacted to produce 1 mole of Fe(OH)3.

Therefore, 9.9×10¯³ mole of KOH will react to produce = (9.9×10¯³ x 1)/3 = 3.3×10¯³ mole of Fe(OH)3.

Finally, we shall convert 3.3×10¯³ mole of Fe(OH)3 to grams. This can be obtained as follow:

Molar mass of Fe(OH)3 = 56 + 3(16 + 1) = 56 + 3(17) = 107 g/mol

Mole of Fe(OH)3 = 3.3×10¯³ mole

Mass of Fe(OH)3 =?

Mole = mass /Molar mass

3.3×10¯³ = Mass of Fe(OH)3 / 107

Cross multiply

Mass of Fe(OH)3 = 3.3×10¯³ x 107

Mass of Fe(OH)3 = 0.3531 ≈ 0.35 g.

Therefore, 0.35 g of Fe(OH)3 was produced from the reaction.

8 0
3 years ago
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