Na2CO3(aq) + CaCl2(aq) — 2 NaCl(aq) + CaCO3(s) Calculate the volume (in mL) of 0.200 M CaCl, needed to produce 2.00 g of CaCO3(s). There is an excess of Na2CO3.
Molar mass of calcium carbonate = 100.09 g/mol

100 mL

Explanation:

Given data:

Mass of CaCO₃ produced = 2.00 g

Molarity of CaCl₂ = 0.200 M

Volume of CaCl₂ needed = ?

Solution:

Chemical equation:

Na₂CO₃  + CaCl₂    →      2NaCl + CaCO₃

First of all we will calculate the number of moles of CaCO₃.

Number of moles = mass/molar mass

Number of moles = 2.00 g / 100.09 g/mol

Number of moles = 0.02 mol

Now we will compare the moles of CaCO₃ and CaCl₂.

              CaCO₃          :            CaCl₂

                   1                :               1

                 0.02           :              0.02

Thus, 0.02 moles of CaCl₂ react,

Volume of CaCl₂ reacted:

Molarity = number of moles / volume in L

0.200 M = 0.02 mol / volume in L

Volume in L = 0.02 mol / 0.200 M

Volume in L = 0.1 L

Volume in mL:

0.1 L × 1000 mL/1L

100 mL

the answer for sodium is 0.186

the answer for rubidium is 0.251

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be