Molar equations? How do i figure out a theoretical answer for an experiment
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Molar equations? How do i figure out a theoretical answer for an experiment

[From: ] [author: ] [Date: 12-06-07] [Hit: ]
Each solution was of 25mL. Each solution was left for approximately 4 days.Is it possible to calculate the theoretical results for the experiment. As in, can I calculate how much of the calcium carbonate will dissolve in the solution. If so,......
I conducted an experiment whereby I had different molars of HCl (0.2, 0.4, 0.8, 1.0, 2.0) and following this approximately 1 gram of calcium carbonate marble chips was added to the different solutions. Each solution was of 25mL. Each solution was left for approximately 4 days.

Is it possible to calculate the theoretical results for the experiment. As in, can I calculate how much of the calcium carbonate will dissolve in the solution. If so, could you please explain.

Below are the actual results, which represent how much of the calcium carbonate had dissolved in each solution:
0.2M - 15.15%
0.4M - 39.39%
0.8M - 77.94%
1.0M - 86.60%
2.0M - 87.88%

Thank you!

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The reaction is 2HCl(aq) + CaCO3(s) -----> CaCl2(s) + H2CO3(aq)

Eventually the H2CO3 will dissociate to form CO2(g) and H2O(l)

Excess CaCO3 is used, so HCl is the limiting reactant and determines the amount of CaCO3 that reacts. Each moles of CaCO3 reacted will be 1/2 the moles of HCl.

The molar mass of CaCO3 = 40.08+12.01+48.0= 100 g/mol

1g of CaCO3 is 0.01 mol

In 25 mL of HCl there are 0.025*0.2 = 0.005 mol HCl reacting with 0.0025 mol CaCO3 or 0.250 g CaCO3 = 25.0%
all others are proportional to the molarity of the HCl

0.4 M ----> 50.0%
0.8 M ----> 100%
0.025 L of 1.0 M HCl contains 0.025 moles HCl which reacts with 0.0125 mol CaCO3 which is 1.25 g of CaCO3 which is more than you added, so all of it dissolves. The 2.0 M solution can react with 2.50 g CaCO3, so all of the 1g sample will react.
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