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Do your work on another piece of paper. Show all steps, label all units for full credit. 45ml of 0.155 M Aluminum Chloride reacts with 60 ml of 0.125 M silver sulfate. Complete and Balance the equation and identify the precipitate (1) How many grams of precipitate are formed? If the reaction has a 65% yield, how much precipitate do I get? What is the molarity of the resulting solution? How many grams of solute are in 0.289 liters of a 0.023 M Cu(NO_2) solution? 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? Calculate the final concentration If 2.01 of 0.53 M NaCl, 1.4 I of 0.15M NaCl and 0.6L of water are mixed. Determine the oxidation number of each atom, write the half-reactions, and label them as oxidations or reductions. Cu(s) + HNOl (aq) rightarrow Cu(NO_2)_2(aq) + NO_2(g) + H_2O 95 grams of iron at 98'C (specific heat 0.46 J/g Degree C) is dropped into 150g of water at 35 Degree C (specific heat 4.184 J/g Degree C). What's the equilibrium temperature? Find Delta H^9, for TiO_2(s) + 4HCl(g) rightarrow T.CU(I) + O_2 + 2H_2(g) Given: Ti(s) + O_2 rightarrow riO_2(s) 2HCl(g) rightarrow HH(g) + Cl (g) Delta H = -184.6 kJ Ti (s) + 2Cl_2 (g) rightarrow Delta H = -184.6 kJ 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 1.45 L vessel to 3.68 L against a pressure of 3.64 atmospheres? When potassium chloride reacts with oxygen under the right conditions, potassium chlorate is formed: 2 KCl + 3 O_2 rightarrow 2KClO_2 Given: Delta H_1KCl= -436 kJ/mole and Delta H_1 KCIO_3=391 kJ/mole, determine Delta H.

Explanation / Answer

Q1.

the balance equation

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

balance:

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

the precipitate is AgCl(s)

b

mass of precipitate...

mol of Cl --> MV = 45*0.155 = 6.975

mol of Ag --< 2*MV = 60*0.125 = 7.5

ratio is 1:!

so

Chlorine is limiting

6.975 mmol of AgCl will form

mass = mol*MW = 6.975 * 143.32 = 999.657 g of AgCl will form

c.

if yield is only 65% then, 0.65*999.657 = 649.77705 g of AgCl will form only

d

M = mol/V

Molarity = [Ag+] + [Al+3] + [SO4-]

Total V = 45+60 =105

so

Molarity = 45*0.155/105 + 0.125*60/105 + leftover of Ag

leftover = (7.5-6.975)/105

Total Molarity = 45*0.155/105 + 0.125*60/105 + (7.5-6.975)/105 = 0.1428M of ions

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