Question

Sodium azide (NaN₃) reacts with sodium nitrate to yield sodium oxide and nitrogen gas. If 41.8 g of NaN₃ reacted with excess sodium nitrate and 11.9 g of nitrogen gas was recovered, what was the percent yield of the reaction?

Answer 1

Stadium a zide (NAN), Sodium nitrate (Nana) The reaction is NON + Nang Nao + Na The balanced reaction is 5 NaNa+ NaNo 3 Nao + 8 Na From the balanced equation 5 moles NaNz produce & moles of Na I mele NaNg = 23+3(14) = 659/mol 1 mole Nz = 289 (: 2X14 = 289) $(65%) of NaNg – 8 (289) of Ni 325g of NaNa - 224g of N If 41.89 of NaNg> - 1224g of Nyn ( 32 59 of Nanez) X(41.89 of Nan) = 28.81 g of Ny ... Theoretical field of Ne gas = 28.81g

Given Actual vield = lling of a. we know that Actual Yield Percent yield = X 100% Theoretical Yield a substitute the values % yield - 11.99 - X 100 % 28-819 = 41.3 % 3 % Yield of Na> 41.3%