Bertie instructs his butler, Jeeves, to make him 40g of calcium oxide (CaO). How much CaCO3 does Jeeves need to find?

The amount of CaCO₃ Jeeves needs to find is approximately 71.392 grams

Explanation:

The given parameters are;

The mass of the CaO required = 40 g

The chemical reaction for the production of CaO from CaCO₃ is given as follows;

CaCO₃(s) → CaO(s) + CO₂(g)

Therefore, 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂,

The molar mass of CaO = 56.0774 g/mol

Number of moles, n = Mass/(Molar mass)

The number of moles of CaO = 40 g/(56.0774 g/mol) ≈ 0.7133 moles of CaO

Therefore, given that 1 mole of CaO is produced from 1 mole of CaCO₃ 0.7133 moles of CaO will be produced from 0.7133 moles of CaCO₃

The molar mass of CaCO₃ = 100.0869 g/mol

Mass = The number of moles × The molar mass

∴ The mass of CaCO₃ Jeeves needs to find = 0.7133 moles × 100.0869 g/mol ≈ 71.392 g

The mass of CaCO₃ Jeeves needs to find ≈ 71.392 grams.

  the   balanced   equation   for   reaction of   ca metal   with   o2 to   produce   cao   is as below

2 ca+ o2→ 2 cao2

  explanation

the reaction is   balanced     since the   number of atoms   in reactant side is   equal to   number of atoms   in   product side.   for   example   they   are 2   atom of   ca in   reactant   side   and   2   atoms of ca   in   the product side.

  the   process   is a   meet the   requirement   of   redox   reaction because :     calcium   is   oxidized   from   oxidation state 0   to 2+   while   oxygen   is reduced from   oxidation   state   0   to 2-