How many milliliters of 0.418 M HCl are needed to react with 52.7 g of CaCO3?
2HCl (aq) + CaCO3 (s) >>> CaCl2 (aq) + CO2 (g) + H20 (l)
I got .00252 mL but its wrong. This is what I did:
52.7 g x (1 mol CaCO3/100.09 CaCO3) x (2 mol HCl/ 1 mol CaCO3)= 1.05 mol HCl
volume= moles/ molarity
volume= 1.05/.418 = .00252 mL
Thanks for showing your work. That makes it easier to find the error.
You're ok to the last step. I would write
L = moles/molarity just so I would remember that the answer is in liters.
Then 1.05/0.418 = 2.52 L or 2520 mL. Check my work for arithmetic errors.
Your calculations seem correct, but there might be a conversion error. Let's go through the steps to confirm the correct answer.
Given:
Mass of CaCO3 = 52.7 g
Molar mass of CaCO3 = 100.09 g/mol
Molarity of HCl = 0.418 M
First, let's find the moles of CaCO3:
moles of CaCO3 = mass / molar mass
moles of CaCO3 = 52.7 g / 100.09 g/mol
moles of CaCO3 = 0.526 mol
Next, using the reaction stoichiometry, we can determine the moles of HCl needed to react with CaCO3. From the balanced equation, we can see that 2 moles of HCl react with 1 mole of CaCO3.
moles of HCl = moles of CaCO3 x (2 moles HCl / 1 mole CaCO3)
moles of HCl = 0.526 mol x (2 mol HCl / 1 mol CaCO3)
moles of HCl = 1.052 mol
Finally, we can calculate the volume of the HCl solution needed using the formula:
volume (mL) = moles / molarity
volume (mL) = 1.052 mol / 0.418 mol/L
volume (mL) = 2.515 mL
So, the correct answer would be 2.515 mL of 0.418 M HCl.
Your initial calculation of 0.00252 mL seems to be incorrect. The mistake might have occurred during unit conversion or calculation, resulting in a lower volume than expected. Double-check your calculations to find the error.