Exam 2 Question 2: Mole, Molecular Weight, and the Ideal Gas Law (25 points Ammo

Question

Exam 2 Question 2: Mole, Molecular Weight, and the Ideal Gas Law (25 points Ammonia (NHa) and gassous hydrogen chloride C) react to form ammonium chlanick according to tme fowowing equation in a reaction, the scientist makes ammonun chloride by mixing 4.21 LOf NHa (2) st 27.0C and 1.02 atm with 535L of HCgat26.0°C and 0.398 atm (1) How marny moles of NHs & and HC ig are there in the reaction? (12) s Tatm Caim) 4.21 L To 08205 0 08205 27:0?: 273 ,15+ 27.0 61 5 5 mel 2) What is the molar ratio betwean the two reactants? (1) 3) "Which is the lmiting reagent, mearing the one that is completely consumed? (2) (4) How many moles of the other reactant are left aher the reaction? (2) (5) How many moles of NHC,s) wil be generated in the end? 14 6) How many grams of NH.CI(s) will be generated in the end? (4) * indicstes chalenging questions

Explanation / Answer

Solution:- (1) The given balanced equation is...

NH3(aq) + HCl(aq) -----> NH4Cl(s)

We can calculate the moles of each using ideal gas law equation, PV = nRT.

For NH3,

n = PV/RT

n = (1.02 atm x 4.21 L)/(0.0821 atm.L.mol-1.K-1 x 300K)

n = 0.174 mol of NH3

For HCl,

n = (0.998 atm x 5.35 L)/(0.0821 atm.L.mol-1.K-1 x 299 K)

n = 0.218 mol HCl

(2) From balanced equation, there is 1:1 mol ratio between two reactants.

(3) NH3 is the limiting reagent as it's moles are less than moles of HCl.

(4) Moles of HCl left after reaction = 0.218 - 0.174 = 0.044 mol

(5) There is 1:1 mol ratio between NH3 and NH4Cl. So, 0.174 moles of NH4Cl are formed.

(6) Molar mass of NH4Cl is 53.49 g/mol. So, grams of it could be calculated as...

0.174 mol x (53.49 g/1mol) = 9.31 g

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