How many mL of 0.528 M HCl are needed to dissolve 9.57 g of CaCO3?

Question

How many mL of 0.528 M HCl are needed to dissolve 9.57 g of CaCO3?

 

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

 

________mL

Answer 1

Hello Hussein

 

The first part of this is answer is to use stoichiometry to convert grams of CaCO₃ to mol HCl.

 

Once we know mol of HCl, you can then calculate the volume using the given molarity of HCl.

 

The stoichiometry process is as follows:

 

g CaCO₃ --> mol CaCO₃ --> mol HCl

 

Step 1. Convert g CaCO₃ to mol CaCO₃ using the molar mass of CaCO₃ = 100 g/mol

 

9.57 g CaCO₃ x 1 mol CaCO₃ / 100 g CaCO₃  =  0.0957 mol CaCO₃

 

Step 2. Convert mol CaCO₃ to mol HCl using the mole ratio from the balanced chemical reaction.  From the reaction you can see that for every 1 mole of CaCO₃ that reacts, you need 2 moles of HCl.

 

0.0957 mol CaCO₃ x 2 mol HCl / 1 mol CaCO₃ = 0.1914 mol HCl ≅ 0.191 mol HCl (to 3 significant figures).

 

Step 3. Use the calculated moles of HCl and the given molarity (0.528 M) to determine the volume of HCl.

 

Molarity = mol / L of solution

 

Rearrange to solve for L of solution, which gives

 

L of solution = mol / Molarity = 0.191 / 0.528 = 0.362 L of HCl

 

Since the question is asking for the volume in mL, use the conversion factor 1 L = 1000 mL to convert L HCl to mL HCl

 

0.362 L x 1000 mL / 1 L = 362 mL