The bonding in iron III nitrite is a. Ionic only b. Covalent only c. Metallic an

Question

The bonding in iron III nitrite is a. Ionic only b. Covalent only c. Metallic and covalent d. Ionic and mettalic e. Ionic and covalent Consider the chemical reaction that occurs when 50 g of dinitrogen tetroxide (N_2O_4, 92.02 g/mol) reacts with 47.5 g of hydrazine (N_2H_4 32.06 g/mol) to Produce nitrogen gas and water. N_2O_4(I) + 2N_2H_4(I) rightarrow 3N_2(g) + 4H_2O(I) If the reactioon Produces 30 g of nitrogen gas, what is the percent yields of this reaction? a. 16.9 % b. 65.7% c. 31.3% d. 48.2% 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. 0.94 b. 1.09 c. 0.39 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. 36.5 liters b. 19.7 liters c. 21.5 liters d. 39.8 liters 16) The density-of gold is 19.3 grams/cm^3. Which equation below represents the conversion of this density to units of nanograms/m^3? a. Density (ng/m^3) = 19.3 times 10^-9 times 10^6 b. Density (ng/m^3) = 19.3 times 10^-9 times 10^6 c. Density (ng/m3) = 19.3 x 10^9 x 10^2 d. Density (ng/m^3) = 19.3 x 10^9 x 10^6 17) Consider the molecular orbital diagram of O_2 with a bond order of 2. How many electrons must be added to the O_2 molecule to reduce the bond order to zero? a. 4 electrons b. 3 electrons c. 2 electrons d. 1 electron

Explanation / Answer

12) The type of bond in Fe(III)nitrate is,

a. Ionic only

13) For the reaction

moles of N2O4 = 50 g/92.02 g/mol = 0.54 mol

moles of N2H4 = 47.5 g/32.06 g/mol = 1.48 mol

Is all of N2O4 is consumed, we would need = 2 x 0.54 = 1.08 mol of N2H4

If all of N2H4 is consumed, we would need = 1.48/2 = 0.74 mol of N2O4

Since moles of N2O4 is less than required, this is the limiting reagent

Theoretical yield of N2 = 3 x 0.54 x 28 = 45.36 g

Percent yield = 30 x 100/45.36 = 66 %

Answer : b. 65..7 %

14) moles of excess reagent left after reaction has gone to completion,

1.48 - 1.08 = 0.40 mol

Answr : c. 0.39 mol

15) At STP, with 54% yield

Volume of N2 formed = 3 x 0.54 mol x 0.54 = 0.875 mol x 22.4 L = 19.7. L

Answer : b. 19.7 L

16) Density ng/m^3

d. density (ng/m^3) = 19.3 x 10^9 x 10^6

17) Number of electrons to be added to reduce bnd order from 2 to zero in O2 would be,

a. 4 electrons

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