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vodomira [7]
1 year ago
13

If the temperature of a gas is increased from 20°C to 35°C, what is the new pressure if the original pressure was 1.2 atm? Assum

e that volume is constant.
Group of answer choices

A. 0.6 atm

B. 1.3 atm

C. 1.0 atm

D. 2.6 atm
Chemistry
1 answer:
Marta_Voda [28]1 year ago
3 0

Answer:

B.) 1.3 atm

Explanation:

To find the new pressure, you need to use Gay-Lussac's Law:

P₁ / T₁ = P₂ / T₂

In this equation, "P₁" and "T₁" represent the initial pressure and temperature. "P₂" and "T₂" represent the final pressure and temperature. After converting the temperatures from Celsius to Kelvin, you can plug the given values into the equation and simplify to find P₂.

P₁ = 1.2 atm                                    P₂ = ? atm

T₁ = 20 °C + 273 = 293 K              T₂ = 35 °C + 273 = 308 K

P₁ / T₁ = P₂ / T₂                                             <----- Gay-Lussac's Law

(1.2 atm) / (293 K) = P₂ / (308 K)                  <----- Insert values

0.0041 = P₂ / (308 K)                                   <----- Simplify left side

1.3 = P₂                                                         <----- Multiply both sides by 308

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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>N₂ + 3H₂ → 2NH₃.</em>

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

  • From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
  • 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.

  • From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).

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