Identify the two principal products of the reaction between ammonia and ethyl propionate
1 answer:
Answer:
Ethyl propionate reacts with ammonia it gives ethanamide and ethanol.
Explanation:
When an ester reacts with ammonia to give acid amide and an alcohol. This can be expressed in the form of chemical reaction :
Ethyl propionate is an ethyl ester of propionic acid.
Ethyl propionate =
So, when ethyl propionate reacts with ammonia it gives ethanamide and ethanol.
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Answer:
Explanation:
You don't give the reaction, but we can get by just by balancing atoms of Na.
We know we will need the partially balanced equation with masses, moles, and molar masses, so let’s gather all the information in one place.
M_r: 142.04
2NaOH + … ⟶ Na₂SO₄ + …
n/mol: 0.75
1. Use the molar ratio of Na₂SO₄ to NaOH to calculate the moles of NaF.
Moles of Na₂SO₄ = 0.75 mol NaOH × (1 mol Na₂SO₄/2 mol NaOH
= 0.375 mol Na₂SO₄
2. Use the molar mass of Na₂SO₄ to calculate the mass of Na₂SO₄.
Mass of Na₂SO₄ = 0.375 mol Na₂SO₄ × (142.04 g Na₂SO₄/1 mol Na₂SO₄) = 53 g Na₂SO₄
The reaction produces of Na₂SO₄.
(a) 33.6 L of oxygen would be produced.
(b) 106 grams of would be needed
1 mole of gas = 22.4 L
(a) From the equation, 2 moles of produce 3 moles of
. 1 mole of
will, therefore, produce 1.5 moles of
.
1.5 moles of oxygen = 22.4 x 1.5 = 33.6 L
(b) 22.4 L of is produced at STP. This means that 1 mole of the gas is produced.
From the equation, 1 mole of requires 1 mole of
.
Molar mass of = (23x2)+ (12)+(16x3) = 106 g/mol
Mass of 1 mole = 1 x 106 = 106 grams
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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