Equation is:
Ca(NO3)2 + Na3PO4 —> NaNO3 + Ca3(PO4)2
Balance it out
3Ca(NO3)2 + 2Na3PO4 —> 6NaNO3 + Ca3(PO4)2
You have 96.1g of Calcium Nitrate
Moles = Mass/Molar Mass
Molar mass of Calcium Nitrate is 164.088 g/mol
96.1g / 164.088 g/mol =0.586 moles
Mole ratio is 3:1
Therefore 0.586/3 moles of calcium phosphate
Moles x molar mass = mass
0.586/3 x 310.18 g/mol = 60.6~g roughly as theoretical yield
My best guess has to be 1:3
Answer is Manganese(ii) chloride
To answer this, you need to know the charges for both Mn and Cl
Mn is in the transition metal category and is unique with having 3 levels of charges
manganese - MN
Manganese(ii) - Mn^2+
Manganese(iii) - Mn^3+
Cl is a nonmetal with the standard charge of Cl^-1
Since you have two charges that don't cancel out, they get switched.
So Cl would receive a positive 2. Because it was a positive 2, we can assume that it starts with Manganese(ii). Then we add chloride to the end of it as it ends with Cl.
