When water was added to a 4.00 gram mixture of potassium oxalate hydrate (molar mass 184.24 g/mol) and calcium hydrate shown below , 1.20 g of calcium oxalate hydrate (146.12 g/mol molar mass) was recovered. if the mole: mole ratio of potassium oxalate to calcium oxalate is 1: 1 , what percentage of the 4g-mixture is potassium oxalate

% (COOK)2H2O = 37.826 %

Explanation:

mix: (COOK)2H2O + Ca(OH)2 → CaC2O4 + H2O

∴ mass mix = 4.00 g

∴ mass (CaC2O4)H2O = 1.20 g

∴ Mw (COOK)2H2O = 184.24 g/mol

∴ Mw (CaC2O4)H2O = 146.12 g/mol

∴ r = mol (COOK)2H2O / mol (CaC2O4)H2O = 1

⇒ mass (COOK)2H2O = (1.20 g (CaC2O4)H2O)×(mol (CaC2O4)H2O / 146.12 g (CaC2O4)H2O)×(mol (COOK)2H2O/mol (CaC2O4)H2O)×(184.24 g (COOK)2H2O/mol (COOK)2H2O)

⇒ mass (COOK)2H2O = 1.513 g

⇒ % (COOK)2H2O = ( 1.513 g / 4 g )×100

⇒ % (COOK)2H2O = 37.826 %

  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-