Answer: 1) Maximum mass of ammonia 198.57g
2) The element that would be completely consumed is the N2
3) Mass that would keep unremained, is the one of the excess Reactant, that means the H2 with 3,44g
Explanation:
- In order to calculate the Mass of ammonia , we first check the Equation is actually Balance:
N2(g) + 3H2(g) ⟶2NH3(g)
Both equal amount of atoms side to side.
- Now we verify which reagent is the limiting one by comparing the amount of product formed with each reactant, and the one with the lowest number is the limiting reactant. ( Keep in mind that we use the molecular weight of 28.01 g/mol N2; 2.02 g/mol H2; 17.03g/mol NH3)
Moles of ammonia produced with 163.3g N2(g) ⟶ 163.3g N2(g) x (1mol N2(g)/ 28.01 g N2(g) )x (2 mol NH3(g) /1 mol N2(g)) = 11.66 mol NH3
Moles of ammonia produced with 38.77 g H2⟶ 38.77 g H2 x ( 1mol H2/ 2.02 g H2 ) x (2 mol NH3 /3 mol H2 ) = 12.79 mol NH3
- As we can see the amount of NH3 formed with the N2 is the lowest one , therefore the limiting reactant is the N2 that means, N2 is the element that would be completey consumed, and the maximum mass of ammonia will be produced from it.
- We proceed calculating the maximum mass of NH3 from the 163.3g of N2.
11.66 mol NH3 x (17.03 g NH3 /1mol NH3) = 198.57 g NH3
- In order to estimate the mass of excess reagent, we start by calculating how much H2 reacts with the giving N2:
163.3g N2 x (1mol N2/28.01 g N2) x ( 3 mol H2 / 1 mol N2)x (2.02 g H2/ 1 mol H2) = 35.33 g H2
That means that only 35.33 g H2 will react with 163.3g N2 however we were giving 38.77g of H2, thus, 38.77g - 35.33 g = 3.44g H2 is left
Change the places of 'acts against the motion of an object' and 'causes an object to change speed or direction'
(g solute/g solution)*100 = % mass/mass
30 g / 400 * 100
0,075 * 100
= 7,5% w/w
hope this helps!
The structure of the alkyl bromides used in a malonic ester synthesis of ethyl 2-methyl-4-pentenoate.
Ethyl 2-methyl-4-pentenoate by Malonic ester synthesis.
The alkylation of diethyl malonate or a related ester of malonic acid at the carbon alpha (immediately next) to both carbonyl groups, followed by conversion to a substituted acetic acid, characterizes the chemical reaction known as the malonic ester synthesis.
As a result, it is evident from the structure of ethyl 2-methyl-4-pentenoate that ethyl and methyl bromides are the alkyl bromides employed.
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The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g
<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
- Volume = 250 mL = 250 / 1000 = 0.25 L
- Molarity = 0.3 M
Mole = Molarity x Volume
Mole of Na₂SO₄ = 0.3 × 0.25
Mole of Na₂SO₄ = 0.075 mole
<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
- Molar mass of Na₂SO₄ = 142.05 g/mol
- Mole of Na₂SO₄ = 0.075 mole
Mass = mole × molar mass
Mass of Na₂SO₄ = 0.075 × 142.05
Mass of Na₂SO₄ = 10.65 g
Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.
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