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Save Ans en 14 of 50 > Hydrazine, N, H, reacts with oxygen to form nitrogen...

Question:

Save Ans en 14 of 50 > Hydrazine, N, H, reacts with oxygen to form nitrogen gas and water. N, H. (aq) + 0,(8) N,(g) + 2 H, 0(
Save Ans en 14 of 50 > Hydrazine, N, H, reacts with oxygen to form nitrogen gas and water. N, H. (aq) + 0,(8) N,(g) + 2 H, 0(1) If 3.05 g of N, H, reacts with excess oxygen and produces 0.950 L of N,, at 295 K and 1.00 atm, what is the percent yield of the reaction? percent yield:

Answers

The balanced reaction is as follows:

N2H4(aq) + O2(g) \rightarrow N2(g) + 2H2O(l)

Mass of N2H4 = 3.05 g

Temperature(T) = 295 K

Pressure(P) = 1.00 atm

Determine the moles of N2H4 as follows:

Molar mass of N2H4 = 32.0452 g/mol

= 3.05 g N2H4 x ( 1 mol N2H4 / 32.0452 g N2H4)

= 0.09518 mol N2H4

Use the moles of N2H4 and the mole ratio from the balanced chemical reaction. Determine the moles of N2 as follows:

= 0.09518 mol N2H4 x ( 1 mol N2 / 1 mol N2H4)

= 0.09518 mol N2

The ideal gas law equation is as follows:

PV = nRT

Rearrange the formula for volume as follows:

V = nRT /P

V = (0.09518 mol x 0.08206 L.atm/mol.K x 295 K) / 1 atm

V = 2.304 L

Thus, the theoretical yield of N2 gas is 2.304 L

Determine the percentage yield as follows:

Experimental yield of N2 = 0.950 L

Theoretical yield of N2 = 2.304 L

Percent yield = [ Experimetnal yield / Theoretical yield] x 100 %

Percent yield = [ 0.950 L / 2.304 L] x 100 %

Percent yield = 41.2 %

Thus, the percent yield of the reaction will be 41.2 % [3 S.F].

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