The son of a weaver, Dalton's major contribution to the field of chemistry is his atomic theory proposed in 1803.
Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
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
