A 2.50 g sample of Al(ClO 4 ) 3 is mixed with a 250. mL sample of 0.100 M (NH 4

Question

A 2.50 g sample of Al(ClO4)3 is mixed with a 250. mL sample of 0.100 M (NH4)2CO3.

1.   (2 pts.) Write down the ions present in the two solutions before mixing.

            Al(ClO­4)3 (aq):                                  (NH4)2CO3 (aq):

2.   (2 pts.) Write down the chemical formula of the solid that forms.

3.   (2 pts.) Write the %u201Cmolecular%u201D equation for the process.

4.   (2 pts.) Write the total ionic equation for the process.

5.   (2 pt.) Write the net ionic equation for the process.

6.   (2 pts.) Calculate the mass of solid that forms. [Show your work!]

7. (4 pts.) Calculate the concentrations of all ions present in the solution after mixing. (Assume that the final volume is the sum of the initial volumes.)

      [Show Work!]

8. (4 pts.) [Unrelated to above.] Calculate the volume of 0.375 M HCl required to neutralize 0.500 g of Ca(OH)2. [Show Work! Study the formula of Ca(OH)2 carefully.]

Explanation / Answer

3(NH4)2C2O4 + 2Al(ClO4)3 >> Al2(C2O4)3 + 6NH4ClO4

6 NH4+ (aq) + 3 C2O4 2- (aq) + 2 Al3+ (aq) + 3 ClO4- (aq) >>
Al2(C2O4)3 (s) + 6 NH4+ (aq) + 6 ClO4- (aq)

3 C2O4 2- (aq) + 2 Al3+ (aq) >> Al2(C2O4)3 (s)

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