Sapling Learning ng Leaming Indiana University of Pennsylvania CHEM 100-Fall7-AS

Question

Sapling Learning ng Leaming Indiana University of Pennsylvania CHEM 100-Fall7-ASHE >Activities and Due Dates Homework Atempts Score 11/2/2017 11:55 PM 11.5/14 10/31/2017 01:03 AM Saplng Learning Chlorine gas bn prepared in tn laboratory by he maton of hydrotlonc acid win manga sem) aide You add 33.7 g of MOa to a solution containing 47.9 g of HC What is the lmiting reactant? b) What is the hoecal yield of Ch? tealffn yeed ofthe reodon is 81.1%, what is the aduai yield ofchlorine? 12 Search the web and Windows o e F2 F3 F4 F5 F6 F7 F8 F9 F10

Explanation / Answer

a)

Molar mass of HCl = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = 47.9 g

we have below equation to be used:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(47.9 g)/(36.458 g/mol)

= 1.314 mol

Molar mass of MnO2 = 1*MM(Mn) + 2*MM(O)

= 1*54.94 + 2*16.0

= 86.94 g/mol

mass of MnO2 = 33.7 g

we have below equation to be used:

number of mol of MnO2,

n = mass of MnO2/molar mass of MnO2

=(33.7 g)/(86.94 g/mol)

= 0.3876 mol

we have the Balanced chemical equation as:

4 HCl + MnO2 ---> Cl2 +

4 mol of HCl reacts with 1 mol of MnO2

for 1.3138 mol of HCl, 0.3285 mol of MnO2 is required

But we have 0.3876 mol of MnO2

so, HCl is limiting reagent

Answer: HCl

b)

we will use HCl in further calculation

Molar mass of Cl2 = 70.9 g/mol

From balanced chemical reaction, we see that

when 4 mol of HCl reacts, 1 mol of Cl2 is formed

mol of Cl2 formed = (1/4)* moles of HCl

= (1/4)*1.3138

= 0.3285 mol

we have below equation to be used:

mass of Cl2 = number of mol * molar mass

= 0.3285*70.9

= 23.3 g

Answer: 23.3 g

c)

% yield = actual mass*100/theoretical mass

81.1= actual mass*100/23.3

actual mass=18.9 g

Answer: 18.9 g

Feel free to comment below if you have any doubts or if this answer do not work

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