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Mnenie [13.5K]
3 years ago
11

The solubility of water in diethyl ether has been reported to be 1.468 % by mass.' Assuming that 23.0 mL of diethyl ether were a

llowed to become saturated with water before used and that the 1.2 g magnesium was the limiting reagent, what percentage of the product is expected to be lost due to the water in the solvent if the ether is not anhydrous?
Chemistry
1 answer:
melomori [17]3 years ago
6 0

Explanation:

As it is given that solubility of water in diethyl ether is 1.468 %. This means that in 100 ml saturated solution water present is 1.468 ml.

Hence, amount of diethyl ether present will be calculated as follows.

                          (100ml - 1.468 ml)

                        = 98.532 ml

So, it means that 98.532 ml of diethyl ether can dissolve 1.468 ml of water.

Hence, 23 ml of diethyl ether can dissolve the amount of water will be calculated as follows.

          Amount of water = \frac{1.468 ml \times 23 ml}{98.532 ml}

                                       = 0.3427 ml

Now, when magnesium dissolves in water then the reaction will be as follows.

                Mg + H_{2}O \rightarrow Mg(OH)_{2}

Molar mass of Mg = 24.305 g

Molar mass of H_{2}O = 18 g

Therefore, amount of magnesium present in 0.3427 ml of water is calculated as follows.

           Amount of Mg = \frac{24.305 g \times 0.3427 ml}{18 g}  

                                    = 0.462 g

   

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Calculate the moles of calcium chloride (CaCl2) needed to react in order to produce 85.00 grams of calcium carbonate (CaCO3). us
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Answer:

0.85 mole

Explanation:

Step 1:

The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:

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Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.

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From the balanced equation above,

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Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.

From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3

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