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Nutka1998 [239]
4 years ago
5

Given that the density of the saturated solution is found to be 1.16 g/mL. The molar mass of copper sulfate pentahydrate is 249.

68 g/mol, calculate grams of copper sulfate pentahydrate that will dissolve in 100 g of water at 0oC (show calculations for full credit)
Chemistry
1 answer:
Viktor [21]4 years ago
4 0

Explanation:

Since, it is given that density is 1.16 grams per milliliter and molar mass of copper sulfate pentahydrate is 249.68 g/mol.

Now, as we known that density is the amount of mass present in a unit volume.

Mathematically,         Density = \frac{\text{molar mass}}{volume}

Hence, calculate the volume as follws.

                    Density = \frac{\text{molar mass}}{volume}

                  1.16 g/mL = \frac{249.68 g/mol}{volume}          

                 volume = 215.24 mL

As, 215.24 mL can dissolve in 249.68 g/mol of complex. So, in 100 mL volume amount dissolved will be calculated as follows.

                       \frac{249.68 g/mol}{215.24 mL} \times 100

                    = 116 grams        

Thus, we can conclude that 116 grams of copper sulfate pentahydrate that will dissolve in 100 g of water at 0^{o}C.                              

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Problem PageQuestion Aqueous sulfuric acid will react with solid sodium hydroxide to produce aqueous sodium sulfate and liquid w
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0.72g

Explanation:

Step 1:

We'll begin by writing a balanced equation for the reaction. This is illustrated below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Step 2:

Determination of the mass of sulphuric acid (H2SO4) and the mass of sodium hydroxide (NaOH) that reacted from the balanced equation. This is illustrated below:

Molar Mass of H2SO4 = (2x1) + 32 +(16x4) = 2 + 32 + 64 = 98g/mol

Molar Mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Step 3

Determination of the limiting reactant. To do this, we need to know which of the reactant is excess.

Now let us consider using all of the mass of NaOH given to see if there will be left over for H2SO4. This is illustrated below:

From the balanced equation above,

98g of H2SO4 required 80g of NaOH.

Therefore, Xg of H2SO4 will require 1.6g of NaOH i.e

Xg of H2SO4 = (98x1.6)/80

Xg of H2SO4 = 1.96g

Now comparing the mass of H2SO4 that reacted ( i.e 1.96g) and the mass of H2SO4 given ( i.e 2.94g), we can see clearly that there are left over ( i.e 2.94 - 1.96 = 0.98g) of H2SO4. Therefore, H2SO4 is the excess reactant and NaOH is the limiting reactant.

Step 4:

Determination of the mass of water produced from the reaction. This is illustrated below:

The balanced equation for the reaction is given below:

H2SO4 + 2NaOH —> Na2SO4 + 2H2O

Molar Mass of H2O = (2x1) + 16 = 2 + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g

From the balanced equation above,

80g of NaOH reacted to produced 36g of H2O.

Therefore, 1.6g of NaOH will react to produce = (1.6 x 36)/80 = 0.72g of H2O.

Therefore, the maximum mass of water (H2O) produced by the chemical reaction of aqueous sulfuric acid with solid sodium hydroxide is 0.72g

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