"""
### Problem Statement
We need to figure out how many grams of precipitate will form when 50.0 mL of 0.150 M calcium nitrate (Ca(NO3)2) solution is mixed with 25.0 mL of 0.200 M sodium sulfate (Na2SO4) solution.
### Steps to Solve the Problem
#### 1. Write the Balanced
Chemical Equation
First, we write the chemical
reaction that happens when
calcium nitrate reacts with
sodium sulfate:
Ca(NO3)2 (aq) + Na2SO4 (aq)
-> CaSO4 (s) + 2NaNO3 (aq)
This equation shows that one
mole of calcium nitrate reacts
with one mole of sodium sulfate
to produce one mole of calcium
sulfate (which is the
precipitate) and two moles of
sodium nitrate.
#### 2. Calculate the Moles of
Each Reactant
We use the formula for moles:
Moles = Molarity * Volume (L)
First, we need to convert
the volume from milliliters
(mL) to liters (L):
50.0 mL = 0.050 L
25.0 mL = 0.025 L
Now, calculate the moles for
each reactant:
For Ca(NO3)2:
Moles of Ca(NO3)2 =
0.150 M * 0.050 L = 0.0075 moles
For Na2SO4:
Moles of Na2SO4 =
0.200 M * 0.025 L = 0.0050 moles
#### 3. Determine the Limiting
Reactant
The limiting reactant is the
one that will be completely
used up first and will
determine how much product
(precipitate) can be formed.
From the balanced equation,
we see that 1 mole of
Ca(NO3)2 reacts with
1 mole of Na2SO4.
Compare the moles of each
reactant:
- Ca(NO3)2: 0.0075 moles
- Na2SO4: 0.0050 moles
Since Na2SO4 has fewer moles,
it is the limiting reactant.
#### 4. Calculate the Moles o
f Precipitate (CaSO4)
The amount of precipitate
formed is determined by the
limiting reactant.
Since the reaction ratio is 1:1,
the moles of CaSO4 will be
equal to the moles of the
limiting reactant, Na2SO4:
Moles of CaSO4 = 0.0050 moles
#### 5. Convert Moles of
Precipitate to Grams
To find out how many grams
of precipitate (CaSO4) will
form, we need the molar mass
of CaSO4.
Molar masses:
- Ca: 40.08 g/mol
- S: 32.06 g/mol
- O4: 4 * 16.00 g/mol
= 64.00 g/mol
Total molar mass of CaSO4:
40.08 + 32.06 + 64.00 = 136.14 g/mol
Now, use the formula to
convert moles to grams:
Mass = Moles * Molar Mass
Mass of CaSO4
= 0.0050 moles * 136.14 g/mol
= 0.6807 grams
### Summary
When 50.0 mL of
0.150 M Ca(NO3)2
solution is mixed with
25.0 mL of 0.200 M Na2SO4
solution, 0.6807 grams of
CaSO4 precipitate will form.
"""