Here we have to draw the four isomers of the compound 3-bromo-4-fluorohexane.
The four isomers of the compound is shown in the figure.
In an organic molecule the chiral -C center is that where four (4) different groups are present. In 3-bromo-4-fluorohexane the 3 and 4 positions are chiral centers. The possible isomers of a molecule can be obtained from the formula 2n. As here 2 chiral centers are present thus number of stereoisomers will be 2×2 = 4.
The four different isomers as shown in the figure are 3R-, 4R-; 3S-, 4S; 3R, 4S and 3S-, 4R- 3-bromo-4-fluorohexane.
In the 3-bromo-4-fluorohexane the functional groups are -Br, C₂H₅, -C₃H₆F and -H for 3-position and -F, -C₂H₅, -C₃H₆ and -H for 4-position respectively.
The priority of the -3 position will be Br > C₃H₆F > C₂H₅ > H and for -4 position F > C₃H₆Br > C₂H₅ > H. If the rotation from the higher priority group to lower is clockwise and anticlockwise then the S- and R- notation are used respectively. However if the -H atom is present at the horizontal position then the notation will be reverse.
Thus the four isomers of the compound is shown.
The third one is correct, not sure abt another one
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
