Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
The number of Ml of C₅H₈ that can be made from 366 ml C₅H₁₂ is 314.7 ml of C₅H₈
<u><em>calculation</em></u>
step 1: write the equation for formation of C₅H₈
C₅H₁₂ → C₅H₈ + 2 H₂
Step 2: find the mass of C₅H₁₂
mass = density × volume
= 0.620 g/ml × 366 ml =226.92 g
Step 3: find moles Of C₅H₁₂
moles = mass÷ molar mass
from periodic table the molar mass of C₅H₁₂ = (12 x5) +( 1 x12) = 72 g/mol
moles = 226.92 g÷ 72 g/mol =3.152 moles
Step 4: use the mole ratio to determine the moles of C₅H₈
C₅H₁₂:C₅H₈ is 1:1 from equation above
Therefore the moles of C₅H₈ is also = 3.152 moles
Step 5: find the mass of C₅H₈
mass = moles x molar mass
from periodic table the molar mass of C₅H₈ = (12 x5) +( 1 x8) = 68 g/mol
= 3.152 moles x 68 g/mol = 214.34 g
Step 6: find Ml of C₅H₈
=mass / density
= 214.34 g/0.681 g/ml = 314.7 ml
It depends on what type of graph you have. The easiest would be using a H-T diagram. Enthalpy of vaporization is the physical change from liquid to vapor. It occurs at a constant pressure and a constant temperature. As shown in the picture, 1 point is drawn on the subcooled liquid, and another point of the saturated vapor isothermal line. Now, the difference between those two points is the value for the enthalpy of vaporization of water.
