Solid lead (IT) s ydrogen sulfide gas. ulfide reacts with aqueous hydrochloric a

Question

Solid lead (IT) s ydrogen sulfide gas. ulfide reacts with aqueous hydrochloric acid to produce solid lead (11) (a) Write the balanced equation for the reaction given above? (b) If 19 grams of lead (IT) sulfide react with 18 grams of hydrochloric acid, how much solid lead (II) chloride can be formed? (c) What is the limiting reagent for the reaction? (d) How many grams of hydrogen sulfide gas is formed? (e) How much of the excess reagent is left over in this reaction? (O If 5.9 grams of solid lead (II) chloride are formed in the reaction, what is the percent yield of this reaction?

Explanation / Answer

a)

Balanced chemical equation is:

PbS + 2 HCl ---> PbCl2 + 2 H2S

b)

Molar mass of PbS,

MM = 1*MM(Pb) + 1*MM(S)

= 1*207.2 + 1*32.07

= 239.27 g/mol

mass(PbS)= 19.0 g

number of mol of PbS,

n = mass of PbS/molar mass of PbS

=(19.0 g)/(239.27 g/mol)

= 7.941*10^-2 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 18.0 g

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(18.0 g)/(36.458 g/mol)

= 0.4937 mol

Balanced chemical equation is:

PbS + 2 HCl ---> PbCl2 + 2 H2S

1 mol of PbS reacts with 2 mol of HCl

for 0.0794 mol of PbS, 0.1588 mol of HCl is required

But we have 0.4937 mol of HCl

so, PbS is limiting reagent

we will use PbS in further calculation

Molar mass of PbCl2,

MM = 1*MM(Pb) + 2*MM(Cl)

= 1*207.2 + 2*35.45

= 278.1 g/mol

According to balanced equation

mol of PbCl2 formed = (1/1)* moles of PbS

= (1/1)*0.0794

= 7.941*10^-2 mol

mass of PbCl2 = number of mol * molar mass

= 7.941*10^-2*2.781*10^2

= 22.1 g

Answer: 22.1 g

c)

PbS is limiting reagent

d)

According to balanced equation

mol of H2S formed = (2/1)* moles of PbS

= (2/1)*0.0794

= 0.1588 mol

mass of H2S = number of mol * molar mass

= 0.1588*34

= 5.4 g

Answer: 5.40 g

e)

According to balanced equation

mol of HCl reacted = (2/1)* moles of PbS

= (2/1)*0.0794

= 0.1588 mol

mol of HCl remaining = mol initially present - mol reacted

mol of HCl remaining = 0.4937 - 0.1588

mol of HCl remaining = 0.3349 mol

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl,

m = number of mol * molar mass

= 0.3349 mol * 36.458 g/mol

= 12.21 g

Answer: 12.2 g of HCl is left

f)

% yield = actual mass*100/theoretical mass

= 5.9*100/22.1

= 26.7 %

Answer: 26.7 %

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