Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3? 2HCl + CaCO3 --> CaCl2 + CO2 + H2O

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃      →    CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of  CaCO₃ with HCl.

                      CaCO₃         :          HCl

                          1               :            2

                      0.087           :         2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

the answer for sodium is 0.186

the answer for rubidium is 0.251

c

explanation: