Extra Credit Assignment when ammonia (NH) reacts with fluorine, the products are

Question

Extra Credit Assignment when ammonia (NH) reacts with fluorine, the products are dinitrogen tetrafluoride 1. (N2F4 and hydrogen fluoride (HF). a. Write the balanced chemical equation b. How many moles of each reactant are needed to produce 4.00 moles of HF? c. How many grams of Fr are needed to prepare 25.5 g of N g 2. What is the volume in liters, of 1.20 g of carbon monoxide gas at 8°C if it has a pressure of 724 mmHg?? 3. What is the strongest type of intermolecular attraction expected in each of each of the following? A) CH4 B) CHCI3 C) NF. D) HF E) All of the molecules above can exhibit hydrogen bonding. 4. Indicate the polarity of each molecule below (polar or nonpolar) A) CO2 B) CH4 C) CHBr3 D) NH3 E) HCN 5. How many moles of chloroethylene (c2H3c) contain 5.47x10 e molecules? 6. A patient received 100. mL of 20% wlv mannitol solution every hour. How many grams of mannitol are given in 1 h? 7. Consider the reaction, 2 Al(OH 3+3 H2SO4 Al2(SO4 3+6 Hao. 8. What is the molarity of 60.0 g of NaoH in 0.250 L of NaoH solution? 9. If you pack a bag of potato chips for a snack on a plane ride, the bag appears to have inflated when you take it out to open. If the initial volume of air in the bag was 250 ml at 760 mmHg, and the plant is pressurized at 650 mmHg, what is the final volume of the bag? 10. Identify the species that is oxidized and the species that is reduced in each reaction. a. 2 Na O2 2 Na20 b. Zn Zno 2 Ag c. CH4 202 CO2 2 H2 d. Fes Al AP* Fee

Explanation / Answer

Q1.

a)

Write the balanced equation

NH3 + F2 = N2F4 + HF

balance N

2NH3 + F2 = N2F4 + HF

balance H

2NH3 + F2 = N2F4 + 6HF

balance F

2NH3 + 5F2 = N2F4 + 6HF

this is now balanced

b)

moles of each reactants required for 4 mol of HF

2NH3 = 6 mol of HF

5F2 =6HF

then, fo r5 mol..

6/4*2 = 3 mol of NH3 required

6/4*5 = 7.5 mol of F2 required

c)

find grams of F2 for:

25.5 g of N2F4

mol of N2F4 = mass/MW = 25.5/104.01 = 0.245 mol of N2F4

then...

2 mol of NH3 = 1 mol of N2F4

0.245 mol of N2F4 requires --> 2*0.245 = 0.49 mol of NH3 required

mass = mol*MW = 0.49*17 = 8.33 g of NH3

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