Chem 1A Review nt added 26.3 g of sodium What is the molarity of the solution? C

Question

Chem 1A Review nt added 26.3 g of sodium What is the molarity of the solution? Calculations You Should Know of sodium nitrate to enough water to give a total volume of 250 Na NO Sqe of 2s00 ml. 20 the concentration of all ions present in solurion present in solui Determine the the mass of calcium phosphate needed to prepare 25 Calculate the 0.0 ml, of a 0.125 M solution. 250.0 x Calculate the mass of zine that will react with 50.0 mL of a 5.89 M solution of HCL · 45 ml of a 356 M ferric nitrate solution reacted with 40.0 mL of a 234 M sodium hydroxide solution. Fe (No),gua on Fe on, + 3hanos a Calculate the mass of solid product produced tA b. Which reactant is in excess and by how many milliliters? 0.39 How many additional mL of the limiting reactant are needed to completely precipitate the insoluble product? c. 91

Explanation / Answer

Page no 91

1 (a) You have the first answer right the one where you are calculating molarity 1.238 M which is the right answer so I will skip

(b) Since the sodium Nitrate solution has a molarity if 1.238 M it will have 1.238 M Na+ and 1.238 M NO3- as

NaNO3 = Na+(aq) + NO3-(aq)

2. Calcium phosphate Ca3(PO4)2 has a MW of 310 g/mol

0.125M in 0.25 L will need 0.125 M x 0.25 L = 0.03125 moles

Which is 0.03125 x 310 = 9.69 g of Calcium phosphate

3. Zinc reacts with HCl as per the equation

Zn + 2 HCl ----> ZnCl2 + H2

50 mL of 5.89 M HCl has

0.050L x 5.89 = 0.2945 moles of HCl

SO we will need 0.2945/2 = 0.1473 moles of Zinc

This is 0.1473 x 65.4 g/mol = 9.63 g of Zinc

4. The equation is

Fe(NO3)3 + 3NaOH ---->Fe(OH)3 + 3NaNO3

65 mL of 0.356 M is 0.065 L x 0.356M = 0.0231 moles of Fe(NO3)3

40 mL of 0.234 M is 0.04 L x 0.234M = 0.00936 moles of NaOH

So the limiting reagent is NaOH

SO the amount of Fe(OH)3 will be 0.00936/3 = 0.00312 moles of Fe(OH)3

which is 0.00312 x 107g/mol = 0.334 g of Fe(OH)3 solid is formed

Fe(NO3)3 is in excess the actual Fe(NO3)3 needed is 0.00312 moles

V x M = moles

V x 0.356M = 0.00312 moles

V = 0.00876 L or 8.76 mL of Fe(NO3)3 is needed

65-8.76 = 56.24 mL of Fe(NO3)3 is excess

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