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3 years ago

if the theoretical yield of a reaction is 26.0 grams and you actually recovered 22.0 grams what is the precent yield

1 answer:

8 0

Percentage Yield = (Actual Yield ÷ Theoretical Yield) × 100

∴ if theoretical yield is 26 g, but only 22.0 is recovered from the reaction,
then Percentage Yield = (22 g ÷ 26 g) × 100
= 84.6 %

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Cl2 + 2OH− → Cl− + ClO− + H2O

Nataly_w [17]

Cl2(g) -------> Cl-(aq) + ClO-(aq)

2e- + Cl2(g) -------> 2Cl-(aq) [reduction]

4OH-(aq) + Cl2(g) -----------> 2ClO-(aq) + 2H2O(l) + 2e- [oxidation]
______________________________________...
2OH-(aq) + Cl2(g) --------> Cl-(aq) + ClO-(aq) + H2O(l)

4 0

3 years ago

How many moles are in 5g of Al? A.0.185 B.18.5 C.5.4 D.135

Lapatulllka [165]

I got 134.91 but if you round it you’ll get 135

3 0

3 years ago

Read 2 more answers

How many grams of aluminum chloride are produced when 5.96 grams of aluminum are reacted with excess chlorine gas? Start with a

Vadim26 [7]

Answer:

29.47 g of AlCl₃.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2Al + 3Cl₂ —> 2AlCl₃

Next, we shall determine the mass of Al that reacted and the mass of AlCl₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 2 × 27 = 54 g

Molar mass of AlCl₃ = 27 + (35.5× 3)

= 27 + 106.5

= 133.5 g/mol

Mass of AlCl₃ from the balanced equation = 2 × 133.5 = 267 g

SUMMARY:

From the balanced equation above,

54 g of Al reacted to produce 267 g of AlCl₃.

Finally, we shall determine the mass of AlCl₃ produced by the reaction of 5.96 g of Al. This can be obtained as follow:

From the balanced equation above,

54 g of Al reacted to produce 267 g of AlCl₃.

Therefore, 5.96 g of Al will react to produce = (5.96 × 267)/54 = 29.47 g of AlCl₃.

Thus, 29.47 g of AlCl₃ were obtained from the reaction.

3 0

2 years ago

What is the purpose of the following in the experiment of Synthesis of Methyl Salicylate

Aneli [31]

Answer:

1. Phosphoric Acid : Catalyst

2. Methyl Anthranilate : Reactive

3. Sodium Nitrite : Reactive

4. Diethyl Ether : Solvent and reactant

5. Nitrogen : Sub-product

Explanation:

The phosphoric acid is used as a catalyst for the reaction, the methyl anthranilate will react with the sodium nitrite to produce methyl salicylate, along with the diethyl ether and the nitrogen is a sub-product of the reaction.

7 0

3 years ago

5.(a) Give the names and symbols of respective elements. (5)

Leya [2.2K]

Answer:

Elements:

Platinum - Pt

Silicon - Si

Lead - Pb

Hydrogen- H

Silver - Ag

Symbols:

Pr - Praseodymium

K - Potassium

Br - Bromine

Mg - Magnesium

Hg - Mercury

6 0

2 years ago