Question

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.

If you are able to figure that one out and still have extra time, how many grams of the solid product is amde when 3.45 mL of .553M Ca(NO3)2 reacts with excess 1.239 (NH4)3PO4 soln?

Answer 1

Please refer to the solving processes below...

Explanation:

`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`