Please check my mass of reactants?

Question

H₂S(aq) Pb(NO₃)₂(aq) ---> PbS(s) 2HNO₃(aq)

Mass of lead(II) sulfate precipitated

n(Pb(NO₃)₂) = 0.0157

Pb(NO₃)₂ : Pbs so 1 : 1

so: 0.0157 x (207.19 32.064)

= 3.76g

Is this correct?

The 2nd question is finding the mass of hydrogen sulfide used in the reaction. Do I need to do 1 : 1 : 1 fore mole ratio? or Just use Hydrogen sulfate to Lead Sulfide

Answer 1

H2S  &  Pb(NO3)2 ---> PbS  &  2HNO3

find the Mass of lead(II) sulfide precipitated from 0.0157 moles of Pb(NO3)2

Pb(NO3)2 ---> PbS   , 1 : 1  , so:

0.0157 moles of Pb(NO3)2 = 0.0157 moles of PbS

0.0157 moles of PbS @  239.28 g/mol

= 3.76g

Yes you are right

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The 2nd question is finding the mass of hydrogen sulfide used in the reaction.

both ways will work, i'd stick with the way you did the first problem

1 H2S  & 1 Pb(NO3)2 --->1 PbS  &  2HNO3

1 H2S  & 1 Pb(NO3)2 --->    , 1 : 1  , so:

0.0157 moles of Pb(NO3)2 = 0.0157 moles of H2S

0.0157 moles of H2S @  34.08 g/mol

your answer is: 0.54g H2S