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sergejj [24]
3 years ago
11

Hydrazine (N2H4), a rocket fuel , reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the

equation:
N2H4(l) + O2(g) → N2(g) + 2H2O(g)

At STP, if 4.20L of O2 reacts with N2H4, how many liters of water vapor will be produced?

A)
2.10L H2O(g)


B)
2.67L H2O(g)


C)
5.33L H2O(g)


D)
8.40L H2O(g)
Chemistry
1 answer:
lilavasa [31]3 years ago
3 0

Answer:

D)  8.40 L H₂O(g).

Explanation:

  • The balanced equation for the mentioned reaction is:

<em>2N₂H₄(l) + O₂(g) → N₂(g) + 2H₂O(g), </em>

It is clear that 2.0 moles of N₂H₄ react with 1.0 mole of O₂ to produce 2.0 moles of N₂ and 2.0 moles of H₂O.

  • At STP, 4.20L of O₂ reacts with N₂H₄:

It is known that at STP: every 1.0 mol of any gas occupies 22.4 L.

<u>using cross multiplication: </u>

1.0 mol of O₂ represents → 22.4 L.

??? mol of O₂ represents → 4.2 L.

∴ 4.2 L of O₂ represents = (1.0 mol)(4.2 L)/(22.4 L) = 0.1875 mol.

  • To find the no. of moles of H₂O produced:

Using cross multiplication:

1.0 mol of O₂ produce → 2.0 mol of H₂O, from stichiometry.

0.1875 mol of O₂ produce → ??? mol of H₂O.

∴ The no. of moles of H₂O = (2.0 mol)(0.1875 mol)/(1.0 mol) = 3.75 mol.

  • Again, using cross multiplication:

1.0 mol of H₂O represents → 22.4 L, at STP.

3.75 mol of H₂O represents → ??? L.

<em>∴ The no. of liters of water vapor will be produced </em>= (3.75 mol)(22.4 L)/(1.0 mol) = <em>8.4 L.</em>

<em></em>

<em>So, the right choice is: D)  8.40 L H₂O(g).</em>

<em></em>

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4. How many moles of KBr are found in 3 Liters of 0.4 M solution?
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There are 1.2 moles of KBr found in 3 Liters of 0.4 M solution.

<h3>HOW TO CALCULATE NUMBER OF MOLES?</h3>

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No. of moles = Molarity × volume

According to this question, 3L of a KBr solution are contained in a 0.4M.

no. of moles = 3L × 0.4M = 1.2moles

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A 720. cm^3 vessel contains a mixture of Ar and Xe. If the mass of the gas mixture is 2.966 g at 25.0°C and the pressure is 760.
sleet_krkn [62]

Explanation:

The given data is as follows.

      Pressure (P) = 760 torr = 1 atm

      Volume (V) = 720 cm^{3} = 0.720 L

     Temperature (T) = 25^{o}C = (25 + 273) K = 298 K

Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

   Total atoms present (n) = \frac{PV}{RT}

                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

                                           = 0.0294 mol

Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

               x + y = 0.0294

Also,      40x + 131y = 2.966

             x = 0.0097 mol

              y = (0.0294 - 0.0097)

                = 0.0197 mol

Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

                               = \frac{0.0197}{0.0294}

                              = 0.67

Therefore, the mole fraction of Xe is 0.67.

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