22.0 ammonium chloride decomposes to form hydrochloric acid, and ammonia. at the conclusion of the reaction, the scientist collects 10.24L of ammounia measured at 52 degrees Celsuis and 105.3 kPa. What was the percent yield of the reaction?
Write the equation.
NH4Cl ==> NH3 + HCl
Convert 22.0 (what--you didn't list the units)whatever it is to mols.
mols = grams/molar mass.
Convert mols NH4Cl to mols NH3. They will be equal since 1 mol NH4Cl produces 1 mol NH3. This is the theoretical yield of NH3 in mols.
Use PV=nRT on the NH3 collected.
You have P (convert to atmospheres).
You have V=10.24 L
You have R= constant = 0.08206 L*atm/mol*K.
You have T. Remember to change to Kelvin. K = 273+C.
Solve for n. This is the actual mols NH3 collected.
%yield = [actual mols/theoretical mols]x100
Post your work if you get stuck.
would this be the same thing?
22.0 (1mol/53.492g)=0.411(1L/1L)=0.411(17.034/1 mol)=7.006g - theoretical yield
(1.039atm)(10.24L)/(.0821)(325K)=
0.03987mol(17.034g/1mol)=6.791
and that would be the actual yield?
6.791/7.006 x 100 = 96% yield?
Yes. The method you showed simply involved two more steps. The first one, instead of leaving mols NH3, you multiplied by molar mass NH3 to obtain grams NH3 theoretical. Then after finding the number of mols in PV=nRT, you multiplied again by molar mass of NH3 to obtain grams NH3, actual. Then percent yield is grams NH3 actual/g NH3 theor]x100=
It would have worked the same if you used mols without the conversion to grams.
Yes. The method you showed simply involved two more steps. The first one, instead of leaving mols NH3, you multiplied by molar mass NH3 to obtain grams NH3 theoretical. Then after finding the number of mols in PV=nRT, you multiplied again by molar mass of NH3 to obtain grams NH3, actual. Then percent yield is grams NH3 actual/g NH3 theor]x100=
It would have worked the same if you used mols without the conversion to grams.
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