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meriva
3 years ago
9

A sample of 5.30 mLmL of diethylether (C2H5OC2H5;density=0.7134g/mL)(C2H5OC2H5;density=0.7134g/mL) is introduced into a 6.50 L L

vessel that already contains a mixture of N2N2 and O2O2, whose partial pressures are PN2=0.750 atmPN2=0.750 atm and PO2=0.207 atmPO2=0.207 atm. The temperature is held at 35.0 ∘C∘C, and the diethylether totally evaporates
Chemistry
1 answer:
lilavasa [31]3 years ago
5 0

Answer:

1.16atm

Explanation:

We are going to derive the mass of ether from density

mass=density *volume

Also moles=mass/molecular mass

molar mass C2H5OC2H5 =74.12 g/mole

the density of ether is 0.7134 g/ml

mass C2H5OC2H5 = 5.30 ml x 0.7134 g/ml = 3.78 g

moles C2H5OC2H5 =3.78 g x 1 mole/74.12 g = 0.0509 moles

PV = nRT where P=?; n=0.0509 moles; V=6.50L; R=0.0821 Latm/Kmol; T=35ºC +273 = 308K

P = nRT/V = 0.0509)(0.0821)(308)/6.50

P = 0.198 atm (to 3 significant figures (this is the partial pressure of diethyl ether).  

TOTAL PRESSURE

P1+p2+p3

= 0.198 atm + 0.750 atm + 0.207 atm =1.1550atm

1.16atm(3 significant figures)

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Answer:

1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

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Explanation:

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<em>1) To determine the limiting reactant of the reaction:</em>

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  • This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
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The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.

The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.

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The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).

∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.

0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.

<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>

  • As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
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  • The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
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