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1 year ago

3H2 + blank, reaction arrow, 2NH3

Chemistry

1 answer:

Helga [31]1 year ago

7 0

The complete equation representing the synthesis reaction for the formation of ammonia is: 3 H₂ + N₂ ⇒ 2 NH₃

<h3>What is a synthesis reaction?</h3>

It is a reaction in which 2 or more substances combine to form 1 product.

Let's consider the following incomplete synthesis reaction.

3 H₂ + ____ ⇒ 2 NH₃

According to Lavoisier's law, the mass and the elements must be conserved in a chemical reaction. So, since there is nitrogen to the right, the missing element to the left must be nitrogen. Molecular nitrogen is diatomic.

The complete reaction is:

3 H₂ + N₂ ⇒ 2 NH₃

The complete equation representing the synthesis reaction for the formation of ammonia is: 3 H₂ + N₂ ⇒ 2 NH₃

Learn more about synthesis reaction here: brainly.com/question/16560802

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Substance in mixtures do not change chemically. True or false

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Answer:

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The first compound C6H12 is cyclohexane and the other compound C6H6 is benzene. They are both aromatic compounds. Cyclohexane does not have double bonds in its ring while benzene has three double bonds in its ring. This is why the formula for cyclohexane contains 12 carbon atoms while benzene only has 6.

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3 years ago

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

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Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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I think the awnser to your question is C

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3 years ago

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