Chemistry Problem Solving & Reactions

This chemistry assignment focuses on solving problems related to chemical equations, balancing reactions, solubility, and molecular geometry. Students are asked to write chemical equations for various compounds, balance reactions involving magnesium nitrate, calcium nitrate, and sodium sulfate, determine the molar mass of calcium carbonate, calculate its concentration in solution, and analyze the properties of water, sulfate, and phosphate ions. The assignment also explores the separation of phosphate and sulfate ions using iron(III) nitrate and discusses dilution calculations.

MODULE 1 SCENARIO-BASED PROBLEM
Scene 2
Q1. Chemical equation for:
a) Magnesium nitrate. Mg ( N O3 ) 2
b) Calcium nitrate. Ca ( N O3 ) 2
c) Sodium sulfate. N a2 S O4
Q2. Writing a balanced chemical reaction equations for the following solution
a) magnesium nitrate with sodium sulphate
Mg ( N O3 ) 2 ( aq ) + N a2 S O4 ( aq ) → 2 NaN O3 ( aq ) + MgS O4 (aq)
b) b) calcium nitrate with sodium sulphate
Ca ( N O3 )2 ( aq )+ N a2 S O4 ( aq ) → 2 NaN O3 ( aq ) +Ca S O4 (s)
Q3. Based on the solubility table, calcium sulphate will precipitate out since it is isoluble in
water. When solution of the sodium sulphate is added to magnesium nitrate, the solution
remains colorless. But when sodium sulphate is added to calcium nitrate solutions, a white
precipitate is formed, thus distinguishing beteheen the two solutions (magnesium nitrate
and calcium nitrate).
Q4. Relative atomic mass of Calcium carbonate
a) relative atomic mass of Ca 40.08 amu x 6.022 ×1023 atoms
b) relative atomic mass of C 12.01 amu x 6.022 ×1023 atoms
c) relative atomic mass of O 16.00 amu x 6.022 ×1023 atoms
d) molar mass of CaCO3 100.09 amu
Q5. If one litre of calcium carbonate solution contains 5.34 x 10-3 mol
The concentration is: 5.34 ×10−3 mol
L m=n× MW n=5.34 ×10−3 ; MW =100.09
m=5.34 × 10−3 ×100.09=0.5344806 g

Therefore, Calcium carbonate concentration ¿ 0.5344806 g/ L
Q6. As compare with the standards cited concentration, water with 0.5344806 g/ L can be
considered soft water
Scene 2
Q7. The electron dot structures and Lewis structures of hydroxide, methanal and fluoride are as
follows:
a) Hydroxide OH-
b) Methanal
c) Fluoride
Q8. the Lewis structures of water, sulfate, and phosphate
a) Water
Sulfate

Phosphate
b) The shape are each of these molecules according to VSPER
Water: tetrahedral shape
Sulfate: tetrahedral geometry
Phosphate: tetrahedral molecular geometry
Q9. Lewis structure of the water molecule to indicating which atom(s) have a partial positive
charge and which a partial negative charge.
Q10. Separation of phosphate and sulphate; given sodium chloride, NaCl, iron(III) nitrate,
Fe(NO3)3 calcium nitrate, potassium hydroxide, KOH ammonium fluoride, NH4F.
To the solution containing both anions (phosphate and sulphate), add iron (III)
nitrate to precipitate the phosphate ions. Assuming the solution was sodium phosphate the
precipitation reaction will be:
N a3 PO4 ( aq ) + Fe ( N O3 )3 ( aq ) →3 NaN O3 ( aq )+ FeP O4 ( s)
The phosphate ions will be separated from sulphate ions.
Q11. If the volume of the original sample with concentration 0.25 mg/L is 20 mL, the be the
final volume of solution in order to dilute it to 0.10 mg/L is given as:
Initial molarity, M 1 =0.25 mg/L, initial volume, V 1 = 20 mL
Final molarityM 2 =0.10 mg/L, Final volume V 2=?

M 1 V 1=M 2 V 2V 2= M 1 V 1
M 2
¿ (0.25 mg
L ×20 mL )
0.1 ( mg
L ) =50 mL