Posted by Lexi on Thursday, January 4, 2007 at 5:57pm.
AgNO3+ BaCl2-> AgCl+ Ba(NO3)2
a. balence the equasion
b. How many grams of silver chloride(AgCl) are produced from 5.0g of silver nitrate (AgNO3) reacting with an excess of barium chloride(Ba Cl2)?
c. How Many grams of barium chloride (BaCl2) is necessary to react with 5.0 grams of silver nitrate (AgNO3)?
Can anyone help me with this????
Stoichiometry problems are easy to solve followoing a 4-step process.
Step 1. Write and balance the chemical equation.
2AgNO3 + BaCL2 ==> 2AgCl + Ba(NO3)2
Step 2. Convert what you have (in this case 5.0 g AgNO3) to mols remembering that mols = g/molar mass.
mols AgNO3 = 5.0/170 = 0.294 mols.
Step 3. Convert mols of what you have (in this case mols AgNO3) to mols of what you want (in this case AgCl). Use the coefficients in the balanced equation.
0.294 mols AgNO3 x (2 mols AgCl/2 mols AgNO3) = 0.294 x 2/2 = 0.294 x 1/1 = 0.294 mols AgCl. Note that the fraction is arranged so that the units of AgNO3 cancel and leave units of AgCl. That is how mols of AgNO3 are converted to mols AgCl.
Step 4. Convert mols from step 3 (in this case 0.294 mols AgCl) to grams AgCl. Use the equation from Step 2 rearranged; i.e., mols=grams/molar mass so grams = mols x molar mass and g AgCl = mols AgCl x molar mass AgCl = 0.294 g AgCl x 143 g AgCl/mol AgCl = 4.21 g AgCl.
You will need to redo all of the above because I estimated the molar mass of AgNO3 at 170 and the molar mass of AgCl at 143. Recalculate those values and plug them into the appropriate spots to get an accurate answer. The 4.21 number will be close.
For part 2 of your problem, it is the same procedure. Post your work if you get stuck and need further assistance.
Remember this four-step procedure. It will work many many problems for you. A slight modification is necessary for limiting reagent problems.
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How much barium chloride, BaCl, is necessary to react with the silver nitrate in the problem above?
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