You measured out 1.212g caco3 to create the calcium supplement. how many ml of 1m hcl would you need to completely dissolve the caco3?

-

24.24 mL

Explanation: -

Mass of CaCO₃ = 1.212 g

Molar mass of CaCO₃ = 40 x 1 +12 x 1 + 16 x 3 = 100 g / mol

Number of moles of CaCO₃ =

= 0.01212 mol

The balanced chemical equation for the reaction that occurs when CaCO₃ is added to HCl is

CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂

From the equation we see

1 mol of CaCO₃ reacts with 2 mol of HCl

0.01212 mol of CaCO₃ react with

= 0.02424 mol of HCl

Strength of HCl = 1 M

Volume of HCl required = Number of moles of HCl / strength of HCl

= 0.02424 mol / 1 M

= 0.02424 L

= 0.02424 x 1000 mL

= 24.24 mL

most likely he is cheating on you

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%