Question

How to this question I need help with it please and thank you

Answer 1

First, we write the equation and balance it:

Na2S (s) + Pb(NO3)2 (aq) → 2NaNO3 (aq) + PbS (s) (1)

The reactants:

1) 8.75 g Na2S (s)

The molar mass of Na2S =78.0 g/mol

2) 250 mL of 0.400 mol/L Pb(NO3)2 (aq), we need the mass:

`\begin{gathered} \text{Molarity = }\frac{mass\text{ solute}}{\text{Molar mass of solute x Volume (L)}} \\ M\text{ x Molar mass x V(L)=mass } \\ 0.400\text{ mol/L x 331 g/mol x 0.250 L = 33.1 g} \end{gathered}`

The molar mass of Pb(NO3)2 (aq) = 331 g/mol

Volume = 250 mL = 0.250 L

The mass of Pb(NO3)2 (aq) = 33.1 g

Note: we must determine the limiting reactant first.

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The limiting reactant

"We have 8.75 g Na2S and 33.1 g Pb(NO3)2"

Using stoichiometry and reaction (1),

The molar mass of Na2S =78.0 g/mol

The molar mass of Pb(NO3)2 (aq) = 331 g/mol

78.0 g Na2S ---------- 331 g Pb(NO3)2

8.75 g Na2S ---------- x

x= 37.1 g Pb(NO3)2

Conclusion: For 8.75 g of N2S we need 37.1 g Pb(NO3)2, and according to 2) we only have 33.1 g, we have less than we need, so the limiting reactant is Pb(NO3)2.

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The mass of PbS (s)

We already know the limiting reactant, Pb(NO3)2.

The molar mass of PbS = 239 g/moll

Procedure:

331 g Pb(NO3)2 --------- 239 g PbS

33.1 g Pb(NO3)2--------- y

y= 23.9 g PbS

Answer: Mass of PbS = 23.9 g