The bonding in iron (III) nitrite is a. Ionic only b. Covalent only c. Ionic and

Question

The bonding in iron (III) nitrite is a. Ionic only b. Covalent only c. Ionic and covalent d. Metallic and covalent e. Ionic and metallic For question 13-15 consider the chemical reaction that occurs when 50 g of dinitrogen tetroxide [N_2O_4 92.02 g/mol) reacts with 47.5 f of hydrazine (N_2H_4 32.06 g/mol) to produce nitrogen gas and water. 13. If the reaction produces 30 g of nitrogen gas, what is the percent yield of this reaction? a. 16.9 % b. 31.3 % C. 48.2% d. 65.7% 14) How many moles of excess reagent are left over after the reaction has run to completion if the percent yield of the reaction was 100%? a. 0.39 b. 0.94 c. 1.09 d. 1.48 15) If the reaction is run at STP and the yield is 54% what is the maximum volume of N_2 formed (in liters)? a. 19.7 liters b. 21.5 liters c. 36.5 liters d. 39.8 liters

Explanation / Answer

12) d) metallic and covalent

13)

we can see that

moles = mass / molar mass

so

moles of N204 taken = 50 / 92.02 = 0.54336

moles of N2H4 taken = 47.5 / 32.06 = 1.4816

now

consider the given reaction

N204 + 2N2H4 ---> 3N2 + 4H20

we can see that

moles of N2H4 required = 2 x moles of N204 taken

so

moles of N2H4 required = 2 x 0.54336

moles of N2H4 required = 1.08672

but

1.4816 moles of N2H4 is present

so

N2H4 is in excess and N204 is the limiting reagent

now

moles of N2 produced = 3 x moles of N204 reacted

moles of N2 produced = 3 x 0.54336

moles of N2 produced = 1.63008

now

mass of N2 produced = moles x molar mass

mass of N2 produced = 1.63008 x 28

mass of N2 produced = 45.64224

now

% yield = actual x 100 / theoretical

% yield = 30 x 100 / 45.64224

% yield= 65.7

so

the answer is option d) 65.7 %

14)

excess moles of N2H4 = 1.4816 - 1.08672

excess moles of N2H4 = 0.39488

so

the answer is a) 0.39

15)

% yield = actual x 100 / theoretical

54 = actual x 100 / 45.64224

actual = 24.6468 grams N2

moles = mass / molar mass

moles of N2 = 24.6468 / 28 = 0.88

now

PV = nRT

so

1 x V = 0.88 x 0.0821 x 273

V = 19.73 L

so

the answer is a) 19.7 liters

16)

given

density = 19.3 grams / cm3

we know that

1 gram = 10-9 nanogram (ng)

and

1 cm3 = 10-6 m3

so

density = 19.3 x 10-9 / 10-6

density = 19.3 x 10-9 x 10^6 ng / m3

so

the answer is option b)

17)

d) 4 electrons should be added

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