Here is the chemical reaction - Ca(NO3)2 + (NH4)3PO4 --> Double displacement follows... If you start with 10.25 mL of .553M Ca(NO3)2 soln, how many moles of calcium nitrate do you have in the soln? Also a part 2 in the question.

#A. "1st Condition"#:

Write and balance the equation.

#3Ca(NO_3)_2(aq)+2(NH_4)_3PO_4(aq-Ca_3(PO_4)_2(s)+6NH_4NO_3(aq)#

Given the molarity#(M)# and the volume#(V)# of the solution, the number of moles#(eta)# can be calculated using the formula;

#M=eta/V#

Rearrange the formula to isolate the #eta# and pulg in values as shown below

#eta_(Ca(NO_3)_2)=MxxV#

where:

#M=(0.553mol)/(L)#
#V=10.5cancel(ml)xx(1L)/(1000cancel(ml))=0.0105L#

#eta_(Ca(NO_3)_2)=(0.553mol)/cancel(L)xx0.0105cancel(L)#

#eta_(Ca(NO_3)_2)=0.00657mol#

#B. "2nd Condition"#:

Given the volume#(V)# and the molarity, the number of moles#(eta)# can be calculated as;

#eta_(Ca(NO_3)_2)=MxxV#

where:

#M=(0.553mol)/(L)#
#V=3.45mlxx(1L)/(1000ml)=0.00345L#

#eta_(Ca(NO_3)_2)=(0.553mol)/cancel((L))xx0.00345cancel(L)#

#eta_(Ca(NO_3)_2)=0.00191mol#

Now, find the mole of the solid substance #(Ca_3(PO_4)_2)# in the reaction. Molar conversion is possible with reference to the balanced equation for the molar ratio; i.e.,

#eta_(Ca_3(PO_4)_2)=0.00191cancel(mol Ca(NO_3)_2)xx(1mol Ca_3(PO_4)_2)/(3cancel(molCa(NO_3)_2))#

#eta_(Ca_3(PO_4)_2)=0.000637mol#

Find the molar mass #(Mm)# of the involved compound. Value can be traced in the periodic table; i.e.,

#Mm_(Ca_3(PO_4)_2)=(310.18g)/(mol)#

Now, find the mass of #Ca_3(PO_4)_2# by multiplying its molar mass to its number of moles as computed earlier.

#m_(Ca_3(PO_4)_2)=etaxxMm#

#m_(Ca_3(PO_4)_2)=0.000637cancel(mol)xx(310.18g)/cancel((mol))#

#m_(Ca_3(PO_4)_2)=0.197g#