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July 31, 2015

Homework Help: Chemistry

Posted by chrissy on Sunday, October 15, 2006 at 2:32pm.

Hi! I kind of need help to get started with these four problems. There are more problems like the ones below. It would be great if I had an example of each of the four problems. Thanks!

FeSO4(aq)+ KCL(aq) (for problem 1 and 2)

1)When the following solutions are mixed together, what precipitate (if any) will form?

2)Write the balanced formula equation, complete ionic euation, and net ionic equation, complete ionic equation, and net ionic equation. If no precipitate forms, write "No reaction"

3) What mass of Na2CrO4 is required to precipitate all of the silver ions from 75.0 ml of 0.100 M solution of Ag NO3?

4) What mass of solid aluminum hydroxide can be produced when 50.0 ml of 0.100 M KOH?







FeSO4(aq)+ KCL(aq) (for problem 1 and 2)

1)When the following solutions are mixed together, what precipitate (if any) will form? I suppose you refer to the equation above. The possible products would be K2SO4 and FeCl2. Both of these are soluble in water; therefore, no precipitate will form.

2)Write the balanced formula equation, complete ionic euation, and net ionic equation, complete ionic equation, and net ionic equation. If no precipitate forms, write "No reaction" Since no ppt forms, you should write "No reaction."


3) What mass of Na2CrO4 is required to precipitate all of the silver ions from 75.0 ml of 0.100 M solution of Ag NO3?

4) What mass of solid aluminum hydroxide can be produced when 50.0 ml of 0.100 M KOH?

#3. There is a four-step procedure for doing most of these although the process must be tweaked for some problems.
Step 1. Write the balanced equation.
2AgNO3 + Na2CrO4 ==> Ag2CrO4 + 2NaNO3

Step 2. Convert what you have (in this case AgNO3) to mols.
L x M = mols. 0.0750L x 0.100M = 0.00750 mols.

Step 3. Convert mols of what you have (in this case AgNO3) to mols of what you want (in this case Na2CrO4). To do this, use the coefficients in the balanced chemical equation.
0.00750 mols AgNO3 x (1 mol Na2CrO4/2 mols AgNO3) = 0.003750 mols Na2CrO4. Note the numbers come from the coefficients of the equation. They are placed in the numerator and denominator so that the mols of what you want are on top and the mols of what you have cancel (one in the numerator and one in the denominator). That is the fraction used gives the units you want.

Step 4. Now convert mols in step 3(in this case mols Na2CrO4) to grams. mols x molar mass = g; 0.00375 mols Na2CrO4 x molar mass Na2CrO4 = g Na2CrO4.

For problem 4, I will let you work it by yourself. It is a stoichiometry problem just like 3. Post your work if you get stuck and I can help you through it. You first BIG problem may be writing the equation.


how can preper Ag NO3?


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