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Chemistry

1.a. HNO3 reacts with Fe2S3 according to the following unbalanced chemical equation: HNO3 + Fe2S3 - Fe(NO3)3 H2S Calculate the mass in grams of the excess reagent remaining after the complete reaction of 4.51 g of HNOs with 1.88 g of Fe2S3. 1.b. An aqueous solution of ammonium perchlorate is prepared by dissolving 3.26 g of ammonium perchlorate in 7.89x102 g of water. The density of the solution is 1.60 g mL1. Determine the mass percent of ammonium perchlorate in the solution. 1.c. Determine the mole fraction of ammonium perchlorate in the solution 1.d. An aqueous solution of cobalt (II) nitrate is prepared by dissolving 3.61 g of cobalt(II) nitrate in 5.10x102 g of water. The density of the solution is 1.39 g mL1. Determine the molarity (in mol/L) of cobalt(II) nitrate in the solution.

Explanation / Answer

Ans 1 a

The balanced reaction

6 HNO3 + Fe2S3 = 2 Fe(NO3)3 + 3 H2S

Moles of HNO3 = mass/molecular weight

= 4.51g / 63.01g/mol

= 0.071576 mol

Moles of Fe2S3 = mass/molecular weight

= 1.88g / 207.885g/mol

= 0.009043 mol

From the stoichiometry of the reaction

1 mol Fe2S3 reacts with = 6 mol HNO3

0.009043 mol Fe2S3 reacts with = 0.009043*6

= 0.054258 mol Fe2S3

But we have more moles (0.071576 mol) of HNO3 than required 0.054258 mol.

Limiting reactant = Fe2S3

Excess reactant = HNO3

After completing the reaction

Moles of HNO3 remains = Initial moles - reacted

= 0.071576 - 0.054258

= 0.017318 mol

Mass of HNO3 remains = moles x molecular weight

= 0.017318 mol x 63.01g/mol

= 1.0912 g

Ans 1b

Mass of ammonium perchlorate = 3.26 g

Mass of water = 7.89 x 10^2 g

Total mass of solution = 7.89 x 10^2 + 3.26 = 792.26 g

Density of solution = 1.60 g/mL

Mass% of ammonium perchlorate

= (mass of ammonium perchlorate) *100 / (total mass)

= 3.26*100 / 792.26

= 0.4115 %

Ans 1c

Moles of ammonium perchlorate (NH4ClO4)

= mass of ammonium perchlorate / molecular weight of ammonium perchlorate

= 3.26g / 117.49 g/mol

= 0.027747 mol

Moles of water = mass of water / molecular weight of water

= (7.89 x 10^2 g) / (18.02 g/mol)

= 43.7846 mol

Total Moles = 0.027747 + 43.7846

= 43.81243 mol

Mol fraction of NH4ClO4 = moles of NH4ClO4 / total moles

= 0.027747 / 43.81243

= 6.33 x 10^-4

Ans 1d

Mass of cobalt(II) nitrate = 3.61 g

Moles of cobalt(II) nitrate Co(NO3)2 = mass/molecular weight

= 3.61g / 182.943 g/mol

= 0.0197329 mol

Mass of water = 5.10*10^2 g

Density of solution = 1.39 g/mL

Mass of solution = 3.61 + 5.10*10^2 = 513.61 g

Volume of solution = mass of solution / density of solution

= 513.61 g / 1.39 g/mL

= 369.504 mL x 1L/1000mL

= 0.3695 L

Molarity = moles of cobalt(II) nitrate / volume of solution

= 0.0197329 mol / 0.3695 L

= 0.0534 M

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