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

[nduncan2010]

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

Homework Equations

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

 

[symbolipoint]

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.

 

[Blueshift5]

Hi there,

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It can help us determine the amount of products that will be produced from a given amount of reactants.

In this case, we have the reaction between ammonium sulfide and lead (II) nitrate to form lead (II) sulfide. The balanced equation for this reaction is:

2(NH4)2S + Pb(NO3)2 → 2PbS + 2NH4NO3

To solve this problem, we will use the mole-mole ratio from the balanced equation. This ratio tells us the number of moles of each substance involved in the reaction. We can then use the molar mass of each substance to convert from moles to grams.

Step 1: Calculate the number of moles of ammonium sulfide (NH4)2S.

146.0 g NH4)2S x (1 mol NH4)2S / 68.15 g NH4)2S = 2.14 mol NH4)2S

Step 2: Use the mole-mole ratio to determine the number of moles of lead (II) sulfide (PbS) that will form.

2.14 mol NH4)2S x (1 mol PbS / 2 mol NH4)2S = 1.07 mol PbS

Step 3: Convert the number of moles of PbS to grams.

1.07 mol PbS x 239.27 g PbS / 1 mol PbS = 256.2 g PbS

Therefore, 256.2 grams of lead (II) sulfide will be produced from the combination of 146.0 grams of ammonium sulfide and excess lead (II) nitrate.

I hope this helps! Let me know if you have any further questions.