enChem I Evam Sunmer-1 2025 28) You added to a beaker 1 mole of aqueous Fe(NO,)s

Question

enChem I Evam Sunmer-1 2025 28) You added to a beaker 1 mole of aqueous Fe(NO,)s and 2 moles of aqueous NasCO. What species will be present in the beaker after you stir the mixture? a. iron(lI] ions, nitrate ions, sodium ions, carbonate ions, water b solid iron [IIl) carbonate, sodium lons, nitrate ions, carbonate lons, water c. solid iron (I) carbonate, sodium ions, nitrate ions, iron(lil)ions, water d. solid iron (II) carbonate, sodiums ions, aitrate ions, water e. solid iron (III) carbonate, solid sodium nitrate, water 29) If you want to make 150 mL of a 1.2M nitric acid solution what would you do? a. take 75 mL of a 1.2 M nitric acid stock solution and add 75 ml. ot water. b. take 75 ml of a 0.6 M nitric acid stock solution and add 75 mL of water c. take 75 ml of s 2.4 M nitric acid stock solution and add 75 ml of water d. take 100 ml of a 2.0M nitric acid stock solution and add S0 mL of water e. None of the above

Explanation / Answer

I have solved four problems, post multiple question to get the remaining answers

Q28)

The balanced reaction will be

3Na2CO3 + 2Fe(NO3)3 --> 6NaNO3 + Fe2(CO3)3

Since we are giving it in 2:1 ration, hence Na2CO3 will be the excess reagent, so all the ions will be left

Hence the correct answer is Option B

Q29)

M1V1 = M2V2

(75)M1 = (150)(1.2)

M1 = 2.4M

Hence we must take 75mL of nitric acid of molarity 2.4M and add 75mL of water in the same

Hence the correct answer is Option C

Q30)

Ca + 2HCl ------ CaCl2 + H2

Number of moles of Ca = 100/40 = 2.5 moles

Number of moles of HCl = 100/36.5 = 2.739 moles

1 mole of Ca will react with 2 moles of HCl

Moles of Ca left = 2.5 - 2.739/2 = 1.130 moles

Mass of excess reagent left = 1.130 * 40 = 45.205 grams

Hence the correct answer is Option D

Q31)

Molar mass of Cu = 63.5 gm/mol

Molar mass of Cl = 35.5 gm/mol

Molar mass of water = 18 gm/mol

In case of dried CuClx

63.5/(63.5+35.5x) = 0.473

solving this equation we get, value of x as 2

Hence the dried compound is CuCl2

Now let it contains y moles of water then we can write

63.5/(63.5+2*35.5+18y) = 0.373

18y = 35.74

y = 2

Hence the formula is CuCl2.2H2O

The correct answer is Option D

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