I'm stuck and am hoping you can help. Here is the problem:
You react 7.92 grams of silver nitrate, AgNO3(aq), with an excess of sodium chloride, NaCl(aq). How many grams of silver chloride, AgCl(s), will precipitate from the solution?
Any help would be appreciated.
Most of these stoichiometry problems are worked alike.
Step 1. Write a balanced chemical equation.
AgNO3 + NaCl ==> AgCl + NaNO3.
Step 2. Convert what you have (in this case 7.92 grams AgNO3) to mols remembering that mols = grams/molar mass. The molar mass is approximately 170. You will need to look up the exact value and use it.
mols AgNO3 = 7.92/170 = 0.0466
Step 3. Convert mols of what you have (in this case mols AgNO3) to mols of what you want (in this case AgCl) using the coefficients in the equation from step 1.
0.0466 mol AgNO3 x (1 mol AgCl/1 mol AgNO3) = 0.0466 mol NaCl. Note how the mol AgNO3 in the numerator cancels with the mols in the denominator to leave mols of what you want; i.e., mols AgCl.
Step 4. Now convert mols of what you have into grams using the reverse of step 2; i.e., mols = grams/molar mass, then grams = mols x molar mass.
grams AgCl = mols AgCl x molar mass AgCl.
g AgCl = 0.0466 x 143 (approx) = 6.66 grams.
That's all there is to it. You will need to look up, or calculate, the molar mass of AgNO3 and AgCl. I just rounded those values for convenience. Then you will need to go through the calculations yourself but you should get an answer close to 6.66 g AgaCl.
I hope this helps.
DrBob222 you are amazing!!!!! Seriously I think you just helped me more than you will ever know!
Aint chemistry fun?
When you come to
a. limiting reagent problems
b. problems using volume instead of grams or asking for volume of a product formed instead of grams,
THEN we simply make an adjustment or two to the original four-step process so no extra "learning" is required. This process will stand you in good shape for weight/weight; weight/volume; volume/weight; volume/volume; and limiting reagent problems. My guess is that this four-step process will work about 95% (or more) stoichiometry problems.
how about this...
how many grams of NH3 are needed when 2.96 g of Cr(NO3)3 react according to the reaction: Cr(NO3)3 + NH3 ---> Cr(NH3)6(NO3)3?
hey como hago para contestar las preguntas d emi homework?
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