Question

in my point of veiw, balanced equation- ( 3 mathrm{CaCo}_{3}+2 mathrm{H}_{2} mathrm{Po}_{4} ) gives ( mathrm{Ca}_{3}left(mathrm{Po}_{4}right)_{2}+3 mathrm{Co}_{2}+3 mathrm{H}_{2} mathrm{O} mathrm{so} )
( 210 mathrm{g} ) of ( mathrm{CaCo}_{3} ) and ( 196 mathrm{H}_{3} mathrm{Po}_{4} ) gives ( 310 mathrm{g} ) of
( mathrm{Ca}_{3}left(mathrm{PO}_{4}right) )
According to problem, 50gof ( mathrm{CaCo}_{3} ) and ( 73.5 mathrm{gH}_{3} mathrm{Po}_{4} ) gives ( times mathrm{g} ) of
( mathrm{Ca}_{3}left(mathrm{pO}_{4}right) ) then ( ^{prime} )
( x=(50+73.5) * 310 / 210+196 )
i.e. equal to ( 75.66205 mathrm{g} )

# D-1. 50 g of CaCO3 is allowed to react with 73.5 g of H3PO4. Calculate : (i) Amount of Ca3(PO4)2 formed in moles) (ii) Amount of unreacted reagent (in moles)

Solution