1. According to the following reaction, how many moles of hydrogen gas will be f

Question

1. According to the following reaction, how many moles of hydrogen gas will be formed upon the complete reaction of 27.7 grams of water? water (l) = hydrogen (g) + oxygen (g)

2. For the following reaction, 13.6 grams of Chlorine Gas are allowed to react with 62.3 grams of sodium iodide. Chlorine (g) + sodium iodide (s) = sodium chloride (s) + iodine (s). What is the maximum amount of sodium chloride in g that can be formed? What is the formula for the limiting agent? What is the amount of excess reagent afte the reaction is complete? in (g)

3. For the following reaction, 5.25 grams of phosphorus (P4) are mixed with excess chlorine gas. Reaction yields 16.3 grams of phosphorus trichloride. What are the theoretical yield for phosphorus trichloride and percent yield for the reaction?

4. A 0.4308g sample of pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is 0.9741g. What is the mass percentage of chlorine in the original compound?

5. When 4.389g of a hydrocarbon CxHy were burned in a combustion analysis apparatus, 14.29 g of CO2 and 4.387 g of H2O were produced. In a seperate experiment, the molar mass of the compound was found to be 54.09 g/mol. Determine the empirical and molecular formula of the hydrocarbon.

6. You need to make an aqueous solution of 0.193 M iron (III) sulfate for a lab experiment, using 125 mL volumetric flask. How much solid iron (III) sulfate would you need?

7. In the lab you dissolve 19.3 g of iron (III) sulfate in a volumetric flask and add water for a total volume of 125 mL. What is the molarity of the solution? What is the concentration, in M, of the iron (III) cation? What is the concentration, in M, of the sulfate anion?

8. When 24.7 mL of a 11.0 M nitric acid solution is added to a total volume of 250.0 mL, what is the concentration, in M, of the diluted solution?

9. How many grams of Cu(OH)2 will precipitate when excess KOH solution is added to 67.0 mL of 0.524 M Cu(NO3)2 solution?

10. What volume of a 0.149 M hydroiodic acid solution is required to neutralize 13.1 mL of a 0.154 M potassium hydroxide solution?

11. An aqueous solution of potassium hydroxide is standardized by titration with a 0.154 M solution of hydroiodic acid. If 13.1 mL of base are required to neutralize 13.3 mL of the acid, what is the molarity of the potassium hydroxide solution?

12. A 12.4g sample of an aqueous solution of hydroiodic acid contains an unknown amount of the acid. If 20.8 mL of 0.175 M barium hydroxide are required to neutralize the hydroiodic acid, what is the percent by mass of the hydroiodic acid in the mixture?

13. When 36.2 mL of a 0.162 M iron (III) sulfate solution is combined with 23.4 mL of a 0.365 M iron (III) bromide solution, what is the final concentration, in M, of the iron (III) cation?

Explanation / Answer

1. H2O <==> H2 + 1/2O2

moles of H2 gas formed = 27.7 g/18.015 g/mol = 1.54 mol

2. Cl2 + 2NaI ---> 2NaCl + I2

moles of Cl2 = 13.6 g/70.906 g/mol = 0.192 mol

moles of NaI = 62.3 g/149.89 g/mol = 0.415 mol

If all of Cl2 is consumed we would need = 2 x 0.192 = 0.384 mols of NaI

As number of moles of NaI is greater than needed, Cl2 is the limiting reagent

Maximum mass of NaCl formed = 0.192 mol x 58.44 g/mol = 11.22 g

Mass of excess reagent left = (0.415 - 0.384) mol x 149.89 g/mol = 4.64 g

3. Reaction : P4 + 6Cl2 ---> 4PCl3

moles of P4 = 5.25 g/123.90 g/mol = 0.0424 mol

Theoretical yield of PCl3 = 4 x 0.0424 mol x 137.33 g/mol = 23.29 g

percent yield = (16.3 x 100/23.29) = 70% (or 69.99%)

5. moles of CO2 = moles of C = 14.29 g/44 g/mol = 0.325 mol

moles of H = 2 x moles of H2O = 2 x 4.387 g/18 = 0.487 mol

Divide by smallest factor

C = 0.32/0.32 = 1

H = 0.49/0.32 = 1.5

Empirical formula = C2H3

Molecular formula factor = 54/(24 + 3) = 2

Molecular formula = C4H6

7. moles of Fe2(SO4)3 = 19.3 g/399.88 g/mol = 0.048 mol

Molarity of solution = 0.048 mol/0.125 L = 0.384 M

Molar concentration of Fe(III) = 2 x 0.384 = 0.768 M

Molar concentration of SO4^2- = 3 x 0.384 = 1.152 M

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