How Much Lead(II) Sulfide Is Produced from 146.0 Grams of Ammonium Sulfide?

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The discussion centers on calculating the mass of Lead(II) Sulfide produced from 146.0 grams of Ammonium Sulfide when reacted with excess Lead(II) Nitrate. The balanced chemical equation is 2NH4S + Pb(NO3)2 → 2PbS + 2NH4NO3. Participants emphasize the importance of balancing the equation and using mole ratios to determine the final mass of Lead(II) Sulfide produced. The correct stoichiometric approach is essential for accurate calculations in this reaction.

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stoichiometry. Please help . Thanks!

1. how many grams of Lead(II) Sulfide will result from the combination of 146.0 grams of ammonium sulfide and excess Lead (II) Nitrate



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3. ooohhhkkaayyy .
I cannot even balance the equationn... sooo
this is what i have
2NH4S + Pb (NO3)2 -----> 2PbS + 2NH4 NO3
I CANNOT FIGURE THIS OUT TO SAVE MY LIFEEE
 
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If you understood the FeS problem from your other post (in which you gave a mostly reasonable solution and which was checked and improvement comments responded) then this one is the same idea. Look for the mole ratios from the written reaction, be sure to balance that reaction first; then use formula weights as needed according to the mole ratios.
 

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