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3 years ago

How is the empirical formula of copper chloride hydrate found?

1 answer:

s2008m [1.1K]3 years ago

7 0

CuCl2
IUPAC ID: Copper dichloride, Copper(II) chloride.

Start with the number of grams of each element, given in the problem.
Convert the mass of each element to moles using the molar mass from the periodic table.
Divide each mole value by the smallest number of moles calculated.
Round to the nearest whole number.

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Answer:

22.1g

Explanation:

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

CaO + 2NaCl —> Na2O + CaCl2

Next, we shall determine the mass of CaO that reacted and the mass of Na2O produced from the balanced equation.

This is illustrated below:

Molar mass of CaO = 40 + 16 = 56g/mol

Mass of CaO from the balanced equation = 1 x 56 = 56g

Molar mass of Na2O = (23x2) + 16 = 62g/mol

Mass of Na2O from the balanced equation = 1 x 62 = 62g

From the balanced equation above,

56g of CaO reacted to produce 62g of Na2O.

Finally, we can determine the theoretical yield of Na2O as follow:

From the balanced equation above,

56g of CaO reacted to produce 62g of Na2O.

Therefore, 20g of CaO will react to produce = (20 x 62)/56 = 22.1g of Na2O.

Therefore, the theoretical yield of Na2O is 22.1g

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2 years ago

Under the right conditions aluminum will react with chlorine to produce aluminum chloride.

salantis [7]

Answer:

$m_{Al}=9.42gAl$

Explanation:

Hello there!

In this case, according to the given chemical reaction:

2 Al + 3 Cl2 --> 2 AlCl3

Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

$m_{Al}=11.727LCl_2*\frac{1molCl_2}{22.4LCl_2}*\frac{2molAl}{3molCl_2} *\frac{27.0gAl}{1molAl} \\\\m_{Al}=9.42gAl$

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Answer:

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