General Chemistry 1 Problem Set 2 Do your work on another pi

General Chemistry 1 Problem Set 2 Do your work on another piece of paper. Show all steps, label all units for ful credit 1. (4) 45ml of 0.155 M Aluminum Chloride reacts with 60 mL of 0.125 M silven sulfate. a. 12) Complete and Balance the equation and identify the precipitate b. (3) How many grams of precipitate are formed? c. (05) If the reaction has a 65% yield, how much precipitate do lit? d. (0.5) What is the molarity of the resulting solution? 2. (1) How many grams of solute are in Q.289 Liters of a 0.023 M Cu(NOwz solution? 3. (3) From a 10.0 M NaOH solution, you need to make 250 mL of 0.375M solution How many mL of the NaOH will be required? 4. (2) Calculate the final concentration it 2.0 L of 0.53 M Nacl, 14L of 0.15M NaC and 0.6L of water are mixed. reactions, and CulNOblaq) + label them as oxidations or reductions. Culs) + HNO3(aq) 6. (2) 95 grams of iron at 98C (specific heat 0.46 MgC) is dropped into 150g of water at 35\'C (specific heat 4.184 J/gC), What\'s the equilibrium temperature? (2pt) What is the overall energy change of a system if a chemical reaction transfers 32.1 kJ of heat to the surroundings while it causes the expansion of a 8. 1.45 L vessel to 3.68 L against a pressure of 3.64 atmospheres? (2 pt) When potassium chloride reacts with oxygen under the right conditions, potassium chlorate is formed: 9 Given: * KCI.-436 ki/mol and -KOO-391 k/mol, determine dHne

Solution

1.

(a)

2 AlCl₃ (aq.) + 3 Ag₂SO₄ (aq.) ----------> Al₂(SO₄)₃ (aq.) +6 AgCl (s)

The precipitate is AgCl (white)

(b)

Moles of AlCl₃ = molarity * volume / 1000 = 0.155 * 45 / 1000 = 0.006975 mol

Moles of Ag₂SO₄ = 0.125 * 60 / 1000 = 0.00750 mol

From the balanced equation,

2 mol of AlCl₃ needs 3 mol of Ag₂SO₄

then,

0.006975 mol of AlCl₃ needs 3 * 0.006975 / 2 = 0.0105 mol

But we have 0.00750 mol of Ag₂SO₄ only.

Hence Ag₂SO₄ is limitng reagent.

From the balanced equation,

3 mol of Ag₂SO₄ = 6 mol of AgCl

Then,

0.00750 mol of Ag₂SO₄ = 0.00750 * 6 / 3 = 0.0150 mol of AgCl

Mass of AgCl = moles * molar mass = 0.0150 * 143.3 = 2.15 g.

(c)

Experimental yield = 2.15 * 65 / 100 = 1.40 g.

(d)

Remaining moles of excess reagent = 0.006975 - (0.00750 * 2 / 3) = 0.001975 mol

Volume of solution = 45 + 60 = 105 mL = 0.105 L

Molarity of remaining of AlC3 solution = moles / volume = 0.001975 / 0.105 = 0.0188 M