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11111nata11111 [884]
3 years ago
11

How would you prepare 100 ml of 0.4 M MgSO4 from a stock solution of 2 M MgSO4?

Chemistry
1 answer:
miss Akunina [59]3 years ago
7 0
OK, so to answer this question, you will simply use the molality equation which is as follows:
<span>M1V1 = M2V2 
In the givens you have:
M1 = 2M
V1 is the unknown
M2 = 0.4M
V2 = 100 ml

</span>plug in the givens in the above equation:
<span>2 x V1 = 0.4 x 100 
</span>therefore:
V1 = 20 ml

Based on this: you should take 20 ml of the 2 M solution and make volume exactly 100 ml in a volumetric flask by diluting in water.

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A 1.25 g sample of aluminum is reacted with 3.28 g of copper (II) sulfate. What is the limiting reactant?
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Answer:

d. Copper (II) sulfate

Explanation:

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Mass of Al = 1.25 g

Mass of CuSO₄ = 3.28 g

What is limiting reactant = ?

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Chemical equation:

2Al + 3CuSO₄   →   Al₂ (SO₄)₃ + 3Cu

Number of moles of Al:

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now we will compare the moles of reactant with product.

               Al           :           Al₂ (SO₄)₃

                 2          :             1

               0.05       :          1/2×0.05=0.025 mol

                Al           :            Cu

                 2            :              3

               0.05         :            3/2×0.05 = 0.075 mol

         CuSO₄           :           Al₂ (SO₄)₃

                3             :             1

               0.02         :          1/3×0.02=0.007 mol

         CuSO₄           :            Cu

               3               :              3

               0.02         :              0.02

Less number of moles of reactants are produced by CuSO₄ thus it will act as limiting reactant.

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