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Mnenie [13.5K]
1 year ago
10

A 2.000 g sample of CoCl2.xH2O is dried in an oven. When the anhydrous salt is removed from the oven, its mass is 1.565 g. What

is the value of x?
Chemistry
1 answer:
Mashutka [201]1 year ago
6 0

The value of X is 2

The total amount of sample taken is 2.00 g

The amount of sample left in the oven after drying is 1.565g

The amount of sample lost (mass of water driven out) = Total sample-Anhydrous salt left in the oven

                                 = 2.00 - 1.565

                                = 0.435 grams

The moles of anhydrous salt present in the hydrate = 1.565g/129.83g/mol = 0.01205

The moles of water present in the hydrate = 0.4350g/18.01g/mol = 0.02415

Therefore the ratio of these two are in 1:2 ratio

The complete chemical reaction is   CoCl2.2H20

To know more about the Similar calculation and explanation click here:

brainly.com/question/14967837

#SPJ4

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Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

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Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

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Molarity of solution = 1.86 M

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<u>Answer:</u> The theoretical yield of solid lead comes out to be 5.408 grams.

<u>Explanation:</u>

To calculate the moles, we use the following equation:  

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

  • <u>Moles of Lead nitrate:</u>

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles

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Given mass of aluminium = 2.5 grams

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Putting values in above equation, we get:

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For the given chemical reaction, the equation follows:

2AI(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2AI(NO3)_3(aq.

By Stoichiometry:

3 moles of lead nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead nitrate are produced by = \frac{2}{3}\times 0.0261=0.0174moles of aluminium.

As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.

Lead nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of lead nitrate are produces 3 moles of lead metal.

So, 0.0261 moles of lead nitrate will produce = \frac{3}{3}\times 0.0261=0.0261moles of lead metal.

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Molar mass of lead = 207.2 g/mol

Putting values in above equation, we get:

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