Answer:
The volume of solution in liters required to make a 0.250 M solution from 3.52 moles of solute is 14.08 liters of solution
Explanation:
The question relates to the definition of the concentration of a solution which is the number of moles per liter (1 liter = 1 dm³) of solution
Therefore we have;
The concentration of the intended solution = 0.250 M
Therefore, the number of moles per liter of the required resolution = 0.250 moles
Therefore, the concentration of the required solution = 0.250 moles/liter
The volume in liters of the required solution that will have 3.52 moles of the solute is given as follows;
The required volume of solution = The number of moles of the solute/(The concentration of the solution)
∴ The required volume of solution = 3.52 moles/(0.250 moles/liter) = 14.08 liters
The required volume of solution to make a 0.250 M solution from 3.52 moles of solute = 14.08 liters.
Therefore the number of liters required to make a 0.250 M solution from 3.52 moles of solute = 14.08 liters.
02(g) = 0 kj/mol
<span>CO2 (g) = -393.5 kj/mol </span>
<span>H20(g) = -241.8 kj/mol </span>
<span>H total = -5094 kJ
</span>5094kJ = [8(-393.5) + 9(-241.8)] - [X + 12.5(0)]
<span>-5094 kJ = [-3148 + (-2176.2)] - [x + 0] </span>
<span>-5094 kJ = -5324.2 - x </span>
<span>add -5324.2 to -5094 </span>
<span>to get +230.2 = -x </span>
<span>move the negative to the other side </span>
<span>and you get -230 kj/mol</span>
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
Answer:
true
Explanation:
because if you find the right answer that proves the theory to be correct.
Answer:
-191.7°C
Explanation:
P . V = n . R . T
That's the Ideal Gases Law. It can be useful to solve the question.
We replace data:
2.5 atm . 8 L = 3 mol . 0.082 L.atm/mol.K . T°
(2.5 atm . 8 L) / (3 mol . 0.082 L.atm/mol.K) = T°
T° = 81.3 K
We convert T° from K to C°
81.3K - 273 = -191.7°C
