Question

1. 12.60 milliliters of a 1.25 M KOH solution are required to titrate 20.0 milliliters of a H₃PO₄ solution. What is the molarity of the H₃PO₄ solution?
   3KOH + H₃PO₄ --> K₃PO₄ + 3H₂O  

   		0.513 M
   		2.52 M
   		0.0550 M
   		0.728 M
   		3.04 M
   		0.263 M
   		1.45 M
   		0.817 M

2. Refer to this equation:
   3SCl₂ + 4NaF --> SF₄ + S₂Cl₂ + 4NaCl

   
   64.0 grams SCl₂ is reacted with excess NaF and 12.5 grams of SF₄ is formed.

   What is percent yield SF₄?
   Hint: find theoretical yield
   Hint: percent yield =(actual yield/ theoretical yield) x 100

3. UESTION 19

4. 63.80 grams of a metal that is at temperature 92.50 ^oC is placed in 287.0 mL of water at 21.30 ^oC.   After equilibrium, the final temperature of the water was 26.70 ^oC.   Density of water is 1.00 g/mL. Specific heat of H ₂O is 4.184 J/(g °C). What is the specific heat, c, of the metal?

   		1.54 J/goC
   		1.07 J/goC
   		1.19 J/goC
   		0.670 J/goC
   		2.90 J/goC
   		0.781 J/goC
   		0.105 J/goC
   		0.0711 J/goC

5. How many grams of Fe is needed to react with 7.22 g of sulfuric acid in the following reaction?
   3H₂SO₄ + 2Fe -> Fe₂(SO₄)₃ + 3H₂

   		1.55 g
   		24.9 g
   		42.6 g
   		2.74 g
   		118 g
   		9.21 g
   		4.29 g
   		13.3 g

Answer 1

1) 에 3 KOH + H₂POg - K₂POg + 3H20 Now, 1206 me of 4,25 (M) 6 KOH = 15.75 moot According to above reaction, mole Amount of not = 3 Amount of H * Amount of H₂ POq = 15.75 = 5.25 mmol Now, let, x (M) of H3PO4 is present, come of & (M) 1t3p09 =20x mmol = 5.25 mmol > x = 0.26 25 not (M) . Molarity of H3p0g = 0.263 (M) 3 sel +4 Nat - SFG & Szele & grael 64 g 64 sela- (32 + 71) - = 0.6213 mol sel is reacting with excess Nat. so, limiting reagents sel

0.6213X 1 According to above egnation, excess Nat 3 mol seh can react with and produces mot SFq. 2 0. 6213 mol sela produces mot SFg = 0.2071 mot SF4 .: 012071 mol SFq = (0.2071 X 108.07) g. = 22.3812 gm of Sfq Board ... Theoritieay yields 223812 g of sfg. Actual yields 12.5g of SFq. & Percent yield (mass) = 12,5 x 100 = 55.85% 22.3812 ③ 83.8 g metal at 92.5e is placed in 287 me of water at 21:30. Final Temperature = 260700 let, metal has specific heat & J/g.ee Amount of water = 289 g 7 lgsen) . heat released by metal = heat absorbed by Water - 63-8x x x 1925-26.7) = 287 X 4.189 * ( 26.7 189 + (2 - 21.3) Specific heat of 1420= 4.189]/ge

6484.36 4198.04 x 5) & = 1054 specifie heat of metal = 1.59 g 19.00 4) 3M,504 + 2Fe -> Fez (504), + 3H2 7022 g H2 50g = 7.23 = 0.0736 mol H250g According to above reaction, 3 mol of H₂S0g reacts with a mol Fe. . 0. 0736 mol ito soa wly react with 0.0736x 2 mol Fe = 0.0991 mol re 0.0491 mol Fe = (0.0991x55.85 g : & 2.74g : 2d 2.799 Fe is needed.