Answer:
Van't Hoff factor for AlCl₃ = 3 (Approx)
Explanation:
Given:
Number of observed particular = 1.79 M
Number of theoretical particular = 0.56 M
Find:
Van't Hoff factor for AlCl₃
Computation:
Van't Hoff factor for AlCl₃ = Number of observed particular / Number of theoretical particular
Van't Hoff factor for AlCl₃ = 1.79 M / 0.56 M
Van't Hoff factor for AlCl₃ = 3.19
Van't Hoff factor for AlCl₃ = 3 (Approx)
42700 milliliters would be the answer...
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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
Explanation:
Mass of the organic compound = 200g
Mass of carbon = 83.884g
Mass of hydrogen = 10.486g
Mass of oxygen = 18.640g
The mass of nitrogen = mass of organic compound - (mass of carbon + mass of hydrogen + mass of oxygen)
Mass of nitrogen = 200 - (83.884 + 10.486 + 18.64) = 200 - 113.01
Mass of nitrogen = 86.99g
The empirical formula of a compound is its simplest formula.
It is derived as shown below;
C H O N
Mass 83.884 10.486 18.64 86.99
molar
mass 12 1 16 14
Moles 83.884/12 10.486/1 18.64/16 86.99/14
6.99 10.49 1.17 6.21
Divide
by
lowest 6.99/1.17 10.49/1.17 1.17/1.17 6.21/1.17
6 9 1 5
Empirical formula C₆H₉ON₅
learn more:
Empirical formula brainly.com/question/2790794
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