For questions 13-15, consider the chemical reaction that occurs when 50 g of din

Question

For questions 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 g of hydrazine (N_2H_4, 32.06 g/mol) to produce nitrogen gas and water. N_2O_4(l) + 2N_2H_4(l) rightarrow 3N_2 (g) + 4H_2 O(l) If the reaction produces 30 g of nitrogen gas, what is the percent yield of this reaction? 16.9% 65.7% 31.3% 48.2% 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%? 0.94 1.09 0.39 1.48 If the reaction is run at STI* and the yield is 54%, what is the maximum volume of N_2 formed (in liters)? 36.5 liters 19.7 liters 21.5 liters 39.8 liters The density of gold is 19.3 grams/cm^3. Which equation below represents the of this density to units of nanograms/m^3? Density (ng/m^3) = 19.3 Times 10^-9 Times 10^6 Density (ng/m^3) = 19.3 Times 10^-9 Times 10^2 Density (ng/m^3) = 19.3 Times 10^9 Times 10^2 Density (ng/m^3) = 19.3 Times 10^-9 Times 10^6 Consider the molecular orbital diagram of O_2 with a bond order of 2. How many must be added to the O_2 molecule to reduce the bond order to zero. 4 electrons 3 electrons 2 electrons 1 electron

Explanation / Answer

Solution :-

Q 13-15 )

N2O4 + 2N2H4 --- > 3N2 + 4H2O

Lets first calculate moles of each reactant

Moles = mass / molar mass

Moles of N2O4 = 50 g / 92.02 g per mol = 0.54336 mol

Moles of N2H4 = 47.5 g / 32.06 g per mol = 1.482 mol

Now lets find the moles of N2H4 needed to react with N2O4

0.54336 mol N2O4 * 2 mol N2H4 / 1 mol N2O4 = 1.0867 mol N2H4

Now lets find the moles of N2 that can be formed

0.54336 mol N2O4 * 3 mol N2 / 1 mol N2O4 = 1.63 mol N2

Lets find the theretical yield of the N2

Mass of N2 = moles * molar mass

                      = 1.63 mol * 28.014 g per mol

                      = 45.66 g

Q13) Now lets calculate the percent yield

% yield = ( actual yield / theoretical yield )*100%

              = (30 g / 45.66 g ) *100%

              = 65.70 %

Q14 ) Moles of excess reagent = 1.482 mol N2H4 – 1.0867 mol N2H4 = 0.39 mol N2H4

Q15) if the percent yield of the reaction is 54 %

Then mass of N2 produced = 45.66 g * 54 % / 100 % = 24.65 g

Moles of N2 = 24.65 g / 28.014 g per mol =0.88 mol

Volume of N2 at STP = 0.88 mol * 22.4 L per mol = 19.7 L

So volume is 19.7 L

Q16) 19.3 gram /cm3

Answer is option d

19.3*10^9 ng x 10^6

Q17) to make the bond order zero we have to add 4 electrons

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