Determining volume of an acid needed to react with a mass

[Deluxe489]

1. Determine the volume of 1.50 mol dm–3 of hydrochloric acid that would react with exactly 1.25 g of calcium carbonate.



2. 2HCl(aq) + CaCO3(s) --> CaCl2(aq) + CO2(g) + H2O(1)



3. I tried to do this: (1.25 g CaCO3) / (100 g CaCO3) = 0.0125 moles of CaCO3
2 moles of HCl are needed for every 1 of CaCO3; therefore, 2*0.0125 = 0.0250 moles of HCl
(0.0250 moles of HCl)(36.46 g HCl) / (1.18 g HCl) = 7.73 cm3 HCl
Unfortunately, the correct answer is 16.7 cm 3 HCl. I know it has something to do with the 1.50 mol dm-3, but I don't know how to do the problem.

Thank you for any help.

 

[danago]

Your reasoning up to the point where you conclude that you need 0.025 moles of HCl is good. I am not exactly sure what you have done from there (where did 1.18g come from??)

From the definition of molar concentration, c = n/V, you can find the volume V that you require. You have found that n=0.025 moles of HCl and you have the concentration of HCl, so you can find the volume.

 

[Deluxe489]

Thank you. The density of hydrochloric acid is 1.18 g/cm^3, and I thought I needed to use that. I guess I didn't.

 

[Borek]

The density of hydrochloric acid is 1.18 g/cm^3

This is not a density of hydrochloric acid, this is density of a hydrochloric acid SOLUTION.

 

[DeeNos]

I would first commend you for your efforts in attempting to solve this problem. It is important to carefully analyze and understand the given information before attempting to solve any scientific problem.

In this case, the key information is the concentration of the hydrochloric acid, which is given as 1.50 mol dm-3. This means that for every 1 liter of solution, there is 1.50 moles of HCl present.

To determine the volume of HCl needed to react with 1.25 g of calcium carbonate, we can use the balanced chemical equation provided:

2HCl(aq) + CaCO3(s) --> CaCl2(aq) + CO2(g) + H2O(l)

Based on the equation, we know that for every 2 moles of HCl, 1 mole of CaCO3 is needed. Therefore, we can set up a proportion:

(2 mol HCl / 1 mol CaCO3) = (x mol HCl / 0.0125 mol CaCO3)

Solving for x, we get x = 0.025 mol HCl.

Now, to convert this to volume, we can use the concentration of the HCl solution:

(0.025 mol HCl) / (1.50 mol dm-3 HCl) = 0.0167 dm3 HCl

Finally, we can convert this to cm3 by multiplying by 1000:

0.0167 dm3 HCl * 1000 cm3 / 1 dm3 = 16.7 cm3 HCl

Therefore, the correct answer is indeed 16.7 cm3 of HCl needed to react with 1.25 g of calcium carbonate.

In summary, when solving scientific problems, it is important to carefully analyze the given information and use the appropriate units and equations to arrive at the correct solution. Keep up the good work in your scientific endeavors!